H
y1
IN MEMORIAM
FLORIAN CAJORl
J>?
MODERN GEOMETRY
CAMBRIDGE UNIVERSITY PRESS WAREHOUSE,
C. F. CLAY, Manager.
HonUon: FETTER LANE, E.C.
(f^Umiurgf): 100, PRINCES STREET.
ILdpjig: F. A. BROCKHAUS.
Berlin: A. ASHER AND CO.
i^eijj lotfe : G. P. PUTNAM'S SONS.
aSomfjag mn (Calcutta: MACMILLAN AND CO., Ltd.
[All Bights reserved]
MODERN GEOMETRY
BY
C. GODFREY, M.A.
HEAD MASTER OF THE ROYAL NAVAL COLLEGE, OSBORNE
FORMERLY SENIOR MATHEMATICAL MASTER AT
WINCHESTER COLLEGE
AND
A. W. SIDDONS, M.A,
ASSISTANT MASTER AT HARROW SCHOOL
LATE FELLOW OF JESUS COLLEGE, CAMBRIDGE
CAMBRIDGE
AT THE University Press
1908
PRINTED BY JOHN CLAY, M.A.
AT THE UNIVERSITY PRESS.
CAJORl
PEEFACE
rriHE present volume is a sequel to the Elementary
-^ Geometry written by the same authors.
It covers the schedule of Modern Plane Geometry required
for the Special Examination in Mathematics for the Ordinary
B.A. Degree at Cambridge ; and represents what we take to
be a useful course for any student of Mathematics, whether
he intends to read for Mathematical Honours, or to take
up Physics or Engineering. For those who ultimately make
a special study of Geometry, this book would serve as an
introduction to more advanced treatises.
Our experience tends to shew that boys find no little
difficulty, at the outset, in mastering the ideas of Modern
Plane Geometry; and, especially, in solving the problems
usually set. We have therefore put in a number of quite
easy exercises, arranged to lead by easy stages to the
Scholarship questions at the end of Chapters.
We have to thank Mr H. M. Taylor for permission to
use some of the exercises included in his edition of Euclid.
C. G.
A. W. S.
June, 1908.
918303
CONTENTS.
CHAPTER I.
THE SENSE OF A LINE.
Sense of a line
AB + BC = AC
BA= -AB .
AB + BC + CD + DE+EA = 0
AB=OB-OA .
OA + OB
OC
Sense of an angle
PAGE
1
2
3
3
3
3
5
CHAPTER II.
INFINITY.
Infinity
Point at infinity ....
Line at infinity ....
Circle of infinite radius .
6
7
9
10
Vlll CONTENTS
CHAPTER III.
THE CENTROID.
PAGE
Definition of centroid 11
Theorem 1. If the distance of two points P, Q from a line
XY be yi, y2 (sign being taken into account); and if G be
taken on PQ so that A . PG = ^ . GQ, then the distance of G
from the line is
h-\-k
Theorem 2. If the distances of points Pi, ..., P„ from a line
be yi , ...tyn (sign being taken into account), the distance
of the centroid G from the line is
n
12
13
Theorem 3. If the coordinates of Pi, P2, ..., P„, with respect
to two axes at right angles, be (^1, yi) {x2, 3/2) (•^"s? 3/3)- ••
(^„, yn) the coordinates of the centroid are
n "* n ' ' '
Use of centroid as a method of geometrical proof . . 14
14
CHAPTER IV.
THE TRIANGLE.
Notation 16
Theorem 4. A = |6csinA 17
Theorems. -^ = ^A^ = ^-=2R 18
sm A sin B sin C
Theorem 6. a2 = 62 + c2 — 26ccosA 19
Theorem 7. If a is mid-point of BC, then
AB2+AC2 = 2Aa2 + 2Ba2 (Apollonius' theorem) . . 20
Definition of concurrence, coUinearity 22
Definition of circumcircle, circumcentre ..... 22
CONTENTS IX
PAGE
Theorem 8. The perpendicular bisectors of the sides of a
triangle are concurrent ; and the point of concurrence, S,
is the circumcentre 22
Definition of in-circle, in -centre 23
Theorem 9. The internal bisectors of the angles of a triangle
are concurrent ; and the point of concurrence, I, is the
in-centre 23
Definition of escribed circle or ex-circle, of ex-centre . . 24
Theorem 10. The internal bisector of Z.A and the external
bisectors of Z.s B and C are concurrent ; and the point of
concurrence is the ex-centre li 24
Theorem 11. r = - 25
s
Theorem 12. ri= 26
s—a
Theorem 13. (i) AYi = AZ^ = s. (ii) AY = AZ=5-a.
(iii) YYi = ZZi = a. (iv) BX^^C\ = s-c.
(v) XXi = c-6 27
Definition of median, medial triangle ..... 29
Lemma 1. If y, /3 are the mid-points of AB and AC, then y/3
is parallel to BC and equal to ^BC . . . . 29
Theorem 14. The medians of a triangle are concurrent,
and each median is trisected at the point of concur-
rence, G 30
Centroid of triangle 30
Theorem 15. The three altitudes of a triangle are con-
current . . . • . 31
Definition of orthocentre 31
Definition of pedal triangle 32
Theorem 16. If AH produced meets the circumcircle in X,
then HD = DX 32
Theorem 17. AH = 2Sa . . . . . . . 33,
Theorem 18. The points H, G, S are collinear ; and
HG = 2GS 34
CONTENTS
Theorem 19. A circle whose centre is the mid-point of SH,
and whose radius is ^R, passes through
D, E, F, the feet of the altitudes,
a, ^, y, the mid-points of the sides,
P, Q, R, the mid-points of HA, HB, HC .
Nine-points circle, nine-points centre ....
Theorem 20. If from point P, a point on the circumcircle,
perpendiculars PL, PM, PN be drawn to the sides of a
triangle, then LMN is a straight line (the Simson line)
Exercises on Chapter IV
35
36
37
39
CHAPTER V.
THE THEOKEMS OF CEVA AND MENELAUS.
Lemma 2. If two triangles have the same height, their areas
are to one another in the ratio of their bases . . 46
Theorem 21. (Ceva's theorem.) If the lines joining a point O
to the vertices of a triangle ABC meet the opposite sides
in X, Y, Z, then t;;^; • tt; • ^-^= - 1, the sense of lines
CX AY 3Z.
being taken into account 46
Theorem 22. (Converse of Ceva's theorem.) If points X, Y, Z
are taken on the sides BC, CA, AB of a triangle, such
BX CY AZ
tl^^t r^ • 7^ • s^ = ~ Ij *h®^ '^^^ AX, BY, CZ concurrent . 48
CX AY BZ
Theorem 23. (Menelaus' theorem.) If a straight line
cuts the sides of a triangle ABC in L, M, N, then
BL CM AN , ^ ^, ... 1 . ^ 1
^rr ' -m ' i^rr.= +^, the sense of lines bemg taken uito
CL AM BN °
account .......... 49
Theorem 24. (Converse of Menelaus' theorem.) If points
L, M, N are taken on the sides BC, CA, AB of a triangle,
BL CM AN
such that ^^ . -rj-z . ^xi= +I5 tl^eii ^^^ L, M, N collinear . 50
\.y L> /\ iVI DIN
Exercises on Chapter V .50
CONTENTS XI
CHAPTER VI.
HARMONIC SECTION:
PAGE
DeJ&nition of harmonic section, harmonic range, harmonic
conjugates 53
Definition of cross-ratio ........ 54
Theorem 25. If {AB, CD} = - 1, then ^ + a^ = ^ • • 54
Theorem 26. If AB is divided harmonically at C, D, and if O
is the mid-point of AB, then OC.OD = OB2 ... 55
Definition of pencil, vertex of pencil, transversal ... 56
Theorem 27. If a transversal cuts the four lines of a pencil
at A, C, B, D, and if ACBD is a harmonic range, then any
other transversal will also be divided harmonically . 57
Definition of harmonic pencil 58
Harmonic range with one point at infinity .... 59
Theorem 28. The internal and external bisectors of an angle
are harmonic conjugates with respect to the arms of the
angle 59
Theorem 29. If {AB, CD}= -1 and O is a point outside the
line such that Z.COD is a right angle, then OC, CD are
the bisectors of Z.AOB 59
Exercises on Chapter VI 60
CHAPTER VII.
POLE AND POLAR.
Definition of chord of contact 62
Provisional definition of pole and polar ..... 62
Theorem 30. If the line joining a point T to the centre C of
a circle cuts the chord of contact of T in N and the circle
in A, then CN.CT = CA2 63
Final definition of pole and polar 63
Xll CONTENTS
PAGE
Theorem 31. If a straight line is drawn through any point
to cut a circle, the line is divided harmonically by the
circle, the point, and the polar of the point with respect to
the circle . .■ 64
Theorem 32. If the polar of a point P with respect to a
circle passes through a point Q, then the polar of Q passes
through P 66
Theorem 33. Two tangents are drawn to a circle from a
point A on the polar of a point B ; a harmonic pencil is
formed by the two tangents from A, the polar of B and
the Hne AB 68
Case of circle of infinite radius 69
Exercises on Chapter VII 69
CHAPTER VIII.
SIMILITUDE (pp. 71—75).
CHAPTER IX.
MISCELLANEOUS PROPERTIES OF THE CIRCLE.
Section I. Orthogonal Circles.
Definition of angles at which two curves intersect . . 76
Definition of orthogonal circles 76
Theorem 34. If two circles are orthogonal, a tangent to
either at their point of intersection passes through the
centre of the other ....... 76
Theorem 35. The sum of the squares on the radii of two
orthogonal circles is equal to the square on the distance
between their centres . 77
Theorem 36. Any diameter of a circle which cuts an
orthogonal circle is divided harmonically by the ortho-
gonal circle 77
Section II. The circle op Apollonius.
Theorem 37. If a point P moves so that the ratio of its
distances from two fixed points Q, R is constant, the
locus of P is a circle ....... 78
CONTENTS XIU
PAGE
Section III. Ptolemy's Theorem.
Theorem 38. The sum of the rectangles contained by-
opposite sides of a cyclic quadrilateral is equal to the
rectangle contained by its diagonals .... 80
Theorem 39. The rectangle contained by the diagonals of
a quadrilateral is less than the sum of the rectangles
contained by its opposite sides unless the quadrilateral
is cyclic, in which case it is equal to that sum . . 81
Application of Ptolemy's theorem to trigonometry . . 82
Section IV. Contact Problems 83
Exercises on Chapter IX 84
CHAPTER X.
THE RADICAL AXIS: COAXAL CIRCLES.
Definition of radical axis 87
Theorem 40. The radical axis of two circles is a straight line 87
Theorem 41. The three radical axes of three circles taken in
pairs are concurrent 90
Definition of radical centre 90
Definition of coaxal circles 91
Intersecting coaxal circles . 92
Non-intersecting coaxal circles 93
Limiting points . 94
Theorem 42. With every system of coaxal circles there is
associated another system of coaxal circles, and each circle
of either system cuts every circle of the other system
orthogonally 95
Theorem 43. Of two orthogonal systems of coaxal circles,
one system is of the intersecting type and the other of the
non-intersecting type, and the limiting points of the latter
are the common points of the former .... 97
Exercises on Chapter X 98
XIV CONTENTS
CHAPTER XL
INVERSION.
PAGE
Definition of inverse points ; centre, circle, radius, constant of
inversion . 100
Theorem 44. If a figure is inverted first with one radius of
inversion and then with a different radius, the centre being
the same in both cases, the two inverse figures are similar
and similarly situated, the centre being their centre of
similitude ..... . . .101
Theorem 45. The inverse of a straight line, with regard to a
point on it, is the line itself 102
Theorem 46. The inverse of a straight line, with regard
to a point outside it, is a circle through the centre of
inversion . 102
Theorem 47. The inverse of a circle with regard to a point
on its circumference is a straight line at right angles to
the diameter through the centre of inversion . . . 103
Peaucellier's Cell 104
Theorem 48. The inverse of a circle with regard to a point
not on its circumference is another circle . . . . 105
Theorem 49. Two curves intersect at the same angles as their
inverses 106
Examples of the process of inverting a theorem . . . 107
Inversion in three dimensions 110
Exercises on Chapter XI 110
CHAPTER XII.
ORTHOGONAL PROJECTION.
Definition of orthogonal projection 114
Properties of orthogonal projection 114
Properties of the ellipse 119
CONTENTS XV
CHAPTER XIII.
CROSS-RATIO.
PAOE
Definition of range, base of range . . . . . . 123
Definition of pencil, vertex of pencil 123
Definition of cross-ratio, or anharmonic ratio of range ;
equicross 123
Definition of cross-ratio of pencil 125
Theorem 50. The cross-ratio of a pencil is equal to the cross-
ratio of the range in which any transversal cuts that pencil 126
Theorem 51. If two lines cut a pencil in the ranges ABCD,
A'B'C'D', then {ABCD}= {A'B'C'D'} 127
Theorem 52. If two pencils are subtended by the same range,
then the cross-ratios of the pencils are equal . . . 128
Cross-ratio of pencil of parallel lines 129
Theorem 53. If {ABCD}, {A'B'C'D'} be two equicross ranges,
and if AA', BB', CC be concurrent, then DD' must pass
through the point of concurrence ..... 130
Theorem 54. If two equicross ranges {PXYZ}, {PX'Y'Z'}
have a point P in common, then XX', YY', ZZ' are con-
current 131
Theorem 55. If P {XYZW}, Q {XYZW} be two equicross
pencils, and if X, Y, Z be collinear, then W is on the
line XYZ 132
Theorem 56. If two equicross pencils P{ABCD}, Q{ABCD}
have a ray PQA in common, then BCD are collinear . 133
Cross-ratios are unaltered by projection . . . . . 133
Exercises on Chapter XIII 134
XVI
CONTENTS
CHAPTER XIV.
PKINCIPLE OF DUALITY; COMPLETE QUADRILATERAL
AND QUADRANGLE.
Principle of duality , .
Definition of join of points and meet of lines
Definitions connected with complete quadrilateral, quadrangle
Theorem 57. In a complete quadrilateral, on each diagonal
there is a harmonic range formed by its meets with the
other two diagonals together with two vertices of the
quadrilateral .........
Theorem 58. In a complete quadrangle, through each
diagonal point, there is a harmonic pencil formed by its
joins to the other two diagonal points together with two
sides of the quadrangle .......
Ruler construction for fourth harmonic .....
Self-polar triangle
Theorem 59. If a quadrangle be inscribed in a circle, the
triangle formed by the diagonal points is self-polar with
respect to the circle
Theorem 60. If a quadrilateral be circumscribed about a
circle, the triangle formed by the diagonals is self-polar
with respect to the circle
Triangles in perspective
Theorem 61. (Desargues' theorem.) If two triangles are such
that the lines joining their vertices in pairs are concurrent,
then the intersections of corresponding sides are collinear
Definition of centre, axis of perspective
Note on three-dimensional geometry
Exercises on Chapter XIV
Miscellaneous Exercises
Index
PAGE
136
137
138
139
139
141
142
144
145
146
146
147
148
149
151
161
CHAPTER I.
THE SENSE OF A LINE.
1. Throughout this book the word ' line ' will generally be
used in the sense of ' straight line.'
2. In elementary Geometry, the notation AB as applied to
a straight line has one of two meanings : —
(1) The unlimited straight line defined by, and passing
through, the points A, B.
(2) The limited segment of that line that lies between A
and B.
It is now necessary to explain a third use of the notation.
We may wish to discriminate between the step from A to B,
and the step from B to A. In fact, we may regard AB and BA
as different, AB meaning the step from A to B, and BA the step
from B to A; AB and BA being in different senses. If this idea
is present, it is very usual to draw attention to the fact by
writing a bar over the letters: thus, AB means the step from
A to B.
G. S. M. G. 1
SENSE OF A LINE
A B C
fig. 1.
3. Reverting for a moment to the more elementary idea,
we see that in fig. 1
AB + BC = AC,
and we may interpret this as meaning that the consecutive steps
from A to B, and from B to C, are together equivalent to the
single step from A to C.
It is a natural extension of this if we agree to say that
in fig. 2
AC B
fig. 2.
AB+BC = AC (i),
meaning that, if we step in succession from A to B, and from
B to C, the result is the same as if we had stepped at once from
A to C.
The above is an extension of the idea of addition. The
statement (i) is, in fact, to be regarded as true for all cases, and
as following directly from the extended idea of addition.
SENSE OF A LINE 3
4. As a particular case of the above
AB + BA = 0,
.-. BA=-AB (ii).
If, then, we agree to regard as positive all steps measured
in one sense, we may regard as negative all steps measured in
the opposite sense.
5. Let A, B, C, D, E be any number of points, arranged in
any order along a line.
It follows from the extended idea of addition that
AB + BC + CD + DE = AE.
But AE = - EA,
.*. AB+BC+CD+ DE+ EA-0 (iii).
6. Suppose that O is an origin and A, B any two points
whatever in a line.
I 1 1 _.
BO A
fig. 3.
Then OA + AB = OB,
.-. AB-OB-OA (iv).
7. Let C be the mid-point of AB.
Then 6a + AC = OC,
OB + BC = 00.
But B^ = - AC,
.*. adding, OA + OB = 20C,
. — OA + OB , ,
•• OC-. __ (v).
1—2
4 SENSE OF A LINE
The above results, i — v, are very important and useful :
their value lies in this, that they may be employed with con-
fidence without any reference to tlie figure ; they are true
whatever be the order of the points concerned.
Ex. 1. Verify the truth of the above results, i-v, by taking numerical
instances and placing the points in various orders.
Ex. 2. A, B, C, D are points in any order on a straight fine. Prove
that
AB.CD + AC. DB + AD.BC-0.
Verify by taking numerical instances.
Ex. 3. If AB be divided in C so that m.AC=n.CB, and if O be
any point on the infinite fine AB,
j=~^ m.OA + n. OB
VI + n
Ex. 4. If O, A, B, C be points on a line ; and if P, Q, R be the mid-
points of BC, CA, AB respectively, then
OP . BC + OQ . CA + OR . AB = 0.
Ex. 5. If A, B, C, D be points on a line, and
AC. DB
CB .:ad ~
then
AB.DC ^ ^
^=z — ^= 1 — A.
BC. AD
Ex. 6. If A, B, X, Y are four collinear points, and P is a point on the
same straight line such that PA . PB = PX . PY, show that
PA.BX.BY^PB.AX.AY.
8. Before leaving the subject of the ' sign ' or ' sense ' of
segment of a line, one more remark must be made.
If there be two lines inclined to one another, it is not
possible to compare, as regards sign, segments of the one line
with segments of the other line. In fact, before any such com-
parison is possible we must add ,^/ - 1 to the stock of symbols we
command.
SENSE OF AN ANGLE
The Sense of an Angle.
9. There is a certain analogy (which will be developed later)
between
(a) a point, lying on a certain line, and moving along it, and
ip) a line, passing through a certain point, and rotating
round it.
Just as in case {a) we regarded motion in one sense as
positive and motion in the opposite sense as negative, so in case
(b) we may regard rotation in the one sense as positive and
rotation in the opposite sense as negative.
Thus, if an angle AOB is looked upon as having been swept
out by a radius rotating from OA to OB, we may call it positive;
while, if it is looked at as having been swept out by a radius
rotating from OB to OA, we should call it negative.
When it is convenient to use this idea, we should say that
^AOB==- /.BOA.
CHAPTiiE 11.
INFINITY.
1. There is one exception to the rule that two coplanar
straight lines define a point by their intersection.
This is the case of two parallel straight lines.
There is one exception to the rule that three points define
one circle passing through them.
This is the case of three collinear points.
There is one exception to the rule that a finite straight line
may be divided both internally and externally in a given ratio.
This is the case of the ratio of equality.
These and other exceptions can be removed by means of the
mathematical fiction called ' infinity.'
It will be seen later on that, by means of the concept 'in-
finity ' we are able to state as true without exception an indefinite
number of results which would otherwise have to be stated in a
limited form.
INFINITY
2. Point at infinity on a straight line.
fig. 4.
Let a straight line, always passing through O, start from the
position OP and revolve in a counter-clockwise direction, until it
becomes parallel to the straight line PPi.
In each of its positions, the revolving line cuts the line
PPj in a single point, until the revolving line becomes parallel
to PPi.
When this happens, the statement in black type suddenly
ceases to be true.
The more nearly the revolving line approaches to the parallel,
the more distant does the point of intersection become.
It is found to be convenient to say that the revolving line,
when parallel to PPi, still cuts it; namely, in a point at infinity
on PPj. It will be seen below that these 'mathematical fictions'
— points at infinity — possess many properties in common with
ordinary points.
If the revolving line starts afresh from OP and now revolves
in the clockwise direction, it might be supposed that, when
parallel to PP^, it determines another point at infinity on PPj.
O INFINITY
We shall find, however, that it leads to simpler statements
if we agree to say that this point at infinity is identical with
that obtained formerly.
The reader may object that this is an unreasonable conven-
tion, in that it allows a 'point at infinity ' to be infinitely distant
from itself.
In answer to this objection, it must be explained that we
have not stated that points at infinity enjoy all the properties of
ordinary points.
3. As an illustration of the uniformity of statement obtained
by the coriventions already explained, the following are now given
as true without exception.
(i) Any two coplanar straight lines define one point by
their intersection.
(ii) Two straight lines cannot enclose a space. (If we had
agreed to admit two points at infinity on a straight line, two
parallel straight lines would define two points and would enclose
an infinite space.)
> P B Q
fig. 5.
(iii) If it is required to divide AB in a given ratio, so that,
AP
, — > 1, the problem admits of two solut
ternal division (P) or by external division (Q).
AP
say, ^^> 1, the problem admits of two solutions: either by in-
PB
INFINITY y
If the ratio is gradually altered so that it approaches unity, P
will approach the middle point O, and Q will move off indefinitely
to the right.
When the ratio becomes unity, the internal point of division
is O, and the external point of division is the point at infinity
on AB.
If the ratio had approached unity from below instead of from
above, the internal point of division would have approached O
from the left; and the external point of division would have
moved off indefinitely to the left till, in the limit, it coincided
with the point at infinity, as before.
4. A set of parallel lines cointersect in one point at infinity,
namely the point at infinity belonging to that set. In fact, a set
of parallel lines is a particular case of a set (or pencil) of con-
current lines.
To each set of parallels in a plane, in other words to each
direction, there belongs a point at infinity. If we consider all
possible directions, we have a singly infinite set of points at
infinity; and it may be asked what is the locus of these points.
The locus, apparently, has this property; that every straight
line in the plane cuts it in one point. For a straight line cuts
the locus in the point at infinity on that straight line.
In virtue of the above property, the locus must, itself, be
regarded as a straight line. To call it anything else, e.g. a circle,
would introduce inconsistency of language; and the whole object
of introducing points at infinity is to make mathematical lan-
guage consistent.
The locus of all points at infinity in a plane is, accordingly,
the line at infinity in the plane.
This line has many of the properties of ordinary lines, while
it has other properties that are unfamiliar; e.g. it can be shown
to make any angle whatsoever with itself.
10 INFINITY
5. Limit of a circle of infinite radius.
Suppose that the circle in tig. 6 continually touches the line
DAE in A, while the radius continually increases without limit,
and the centre O recedes to infinity along AF produced.
The circle will flatten out, and the semicircle BAG will clearly
tend to coincide with the infinite line DAE.
But it cannot be supposed that the limit of the circle is simply
DAE ; for a circle is cut by any line in 2 points, while DAE is cut
by any line in 1 point : an essential distinction.
In fact, all the points on the semicircle BFC recede to infinity,
and tend to lie on the line at infinity.
Therefore a circle of infinite radius with centre at infinity
consists of an infinite straight line together with the
straight line at infinity.
Ex. 7. In the limit of figure 6 examine what becomes of the points C,
B and of the tangents EC, DB.
Ex. 8. Find what becomes of the theorem that ' a chord of a circle
subtends equal or supplementary angles at all points of the circumference '
for the case when the circle becomes a finite line plus the line at infinity.
CHAPTER III.
THE CENTROID.
The properties of the centroid are mainly of interest in con-
nection with statics, where they apply to the centre of gravity
of a system of weights. The idea, however, is essentially geo-
metrical; and will, therefore, be developed briefly in this place.
Definition. The centroid of n points in a plane, Pj, P2, P3,
P4, ... P„ is arrived at by the following construction. Bisect PjPg
in A. Divide AP3 in B so that 2AB- BP3. Divide BP4 in C so
that 3BC = CP4; and so forth. The final point obtained by this
process is G, the centroid of P^ ... P^^.*
* The reader will notice that this definition is faulty, inasmuch that
a doubt remains whether we should reach the same point G if we took the
points P in a different order. It is proved below that the point G is
unique.
12
THE CENTROID
Theorem 1.
If the distances of two points p, Q from a line XY be
2/1 J y-2 (sign being taken into account); and if G be taken
on PQ so that A.pg=A;.gq, then the distance of G from
the line is
h + k
P U
fig. 7.
Through P draw PUV || to XY, meeting GL, QN (produced if
necessary) in U, V.
Then, in every case''' (sign being taken into account)
UG : VQ=PG : PG + GQ
= k : k-i-h,
k
:. UG=^ r.VQ.
h-vk
But VQ=NQ-NV=NQ-MP
and LG = UG + MP
= ATI (2/^-2/1) +2/:
h + k
h + k
* This proof is a good instance of the fact explained in Chap. I., that
the attribution of sign to lines makes us, in a measure, independent of the
variety of figures that may be drawn.
THE CENTROID 13
Theorem 2.
If the distances of points Pj, ,.., p„ from a line be
2/1 > ••5 Un (sign being taken into account), the distance
of the centroid G from the line is
The mid-point of Pj P2 is A, and its distance is .
B is taken so that 2AB = BP3,
.'. the distance of B is — =
2 + 1
2/1 + 2/2 + 2/3
3
C is taken so tliat 3BC=:CP4,
*. the distance of C is
3 + 1
2/1 + 2/2 + 2/3 + ^4
4
etc., etc.
Finally the distance of G from the line is
2/1 + 2/2 + 2/3 ••• +2/n
n
14 the centroid
Theorem 3.
If the coordinates of Pj, Pg, ..., P,i, with respect to two
axes at right angles, be {x,, y,) {x.^, y^) (x.„ 3/3) ... (x^, y,,) the
coordinates of the centroid are
n n
This follows immediately from Theorem 2.
Theorem 3 makes it clear, from the symmetry of the expres-
sions for the coordinates of G, that the same centroid would have
been reached if the points P had been taken in any other order.
The centroid is therefore (i) unique, (ii) fixed relative to the
points P.
From the fact that the same centroid is obtained in whatever
order the points are taken, a class of geometrical theorems may be
deduced of which the following is an example.
Example. The medians of a triangle meet in a point,
and each median is trisected at this point.
Consider the centroid of the three points A, B, C. Let BC be
bisected at a, and let G be taken on Aa so that AG = 2Ga. Then
G is the centroid : it lies on the median Aa and trisects it.
Similarly the same point G lies on each of the other medians,
and trisects it.
Hence the medians meet in a point, which trisects each
median.
THE CENTROID 15
Ex. O. ABCD being a quadrilateral, the joins of the mid-points of
AB, CD ; of AC, BD; of AD, BC meet in a point; and each join is bisected
at this point.
Ex. lO. A, B, C, D are four points in a plane. Let the centroids of the
triangles BCD, CDA, DAB, ABC be a, /3, 7, 5 respectively. Then Aa, BjS,
C7, D5 meet in a point ; and are divided in the same ratio at this point.
Ex, 11. Assuming the existence of a centroid in three dimensions,
generalise Exs. 9 and 10 for the case in which ABCD is a tetrahedron.
Ex.12. If G be the centroid of Pi, P2, ... P„, andGMi, GM2, GM3,...
GM„, be the projections of GPi, GP2, GP3, ... GP„ on a line through G ;
then SGIVI=0.
Ex. 13. O being any point, and G the centroid of n points Pj,
P2> ••• Pn>
20P2 = 2GP2+w.OG2.
(Use the extension of Pythagoras' theorem.)
CHAPTER IV.
THE TRIANGLE.
Notatiofi. Special points and quantities will be denoted by
the following letters, in the course of the present chapter
A, B, C vertices of the triangle,
D, E, F feet of the altitudes,
a, yS, y mid-points of the sides,
X, Y, Z points of contact of the in-circle,
a, h, c lengths of the sides,
s semi-perimeter (2s — a -\- h + c)^
R circum-radius,
r in-radius,
^i> ^"2 5 ^'3 ...ex- radii,
A area of the triangle,
S circumcentre,
H orthocentre,
G centroid,
I in-centre,
h, I2, I3 ...ex- centres,
N nine-points centre,
P, Q, R mid-points of HA, HB, HC.
THE TRIANGLE 17
Theorem 4.
A = ^bc sin A.
fig. 9.
Case i. If lA is acute.
Draw CF x to AB.
A = jAB.CF.
But CF = CAsinA,
.'. A^iAB.CAsinA
— \hc sin A.
Similarly A = \ca sin B = \ab sin C.
Case ii. If lA is obtuse.
The proof is left to the reader.
Ex. 14. Prove the above theorem for the case in which z A is obtuse.
Ex. 15. Prove the theorem that the ratio of the areas of similar triangles
is equal to the ratio of the squares on corresponding sides.
Ex. 16. Two sides OP, OR of a variable parallelogram OPQR always
lie along two fixed lines OX, OY ; and Q describes the locus defined by
OP . PQ = constant. Prove that the area of the parallelogram is constant.
Ex. 17. Deduce from Theorem 4 that
O' __ b _ c
sin A ~" sin B ~ sin C *
G. S. M. G. 2
18 THE TRIANGLE
= 2R.
Case i. If the triangle is acute angled.
Join CS.
Produce CS to meet circumcircle in Y.
Join BY.
Since CY is a diameter of the 0 ,
.*. z_ CBY is a rt z. .
Also L BYC = z. BAG,
BC _ a
CY ~ 2R
^ 2R.
sin A
b c
Similarly -r—- = -^—r - 2R.
'' sin B sin C
Case ii. If the triangle is obtuse angled.
The proof of this case is left to the reader.
Ex. 18. Prove Case ii of Theorem 5.
THE TRIANGLE
19
Ex. 10. Prove that
ahc
4A*
Ex. 20. Prove that the circum-radius of an equilateral triangle of side x
is approximately -577^:.
Ex. 21. SAP, PBQ, QCR, RDS are lines bisecting the exterior angles
of a convex quadrilateral ABCD. Prove that
PB . QC . RD . SA = PA . SD . RC . QB.
Ex. 22. Deduce from Theorem 5 the fact that the bisector of the vertical
angle of a triangle divides the base in the ratio of the sides containing the
vertical angle.
Theorem 6.
e^2^52^^2_2^>ccosA.
Case I. If l IK is acute.
a2^52^^2_2c.AF.
But AF = 6cosA,
a^ = 6" + c^ — 26c cos A.
II. 9.
2—2
20 THE TRIANGLE
Case ii. If lPk is obtuse.
a^^b'- + c^ + 2c. AF.
But AF = 6cosCAF
and cos A = - cos CAF,
.*. AF = — i cos A,
.*. a^^b'^ + G'^-2bccosA.
Similarly b"^ = c^ + a^ - 2ca cos B,
c^ = a^ + b^-2abcosC.
Ex. 23. Examine the case z A = 90°.
Theorem 7.
Apollonius'"* Theorem.
If a is mid-point of BC, then
AB2 + AC2=:2Aa2-^2Ba2.
A
II.
Draw AD ± to BC.
Suppose that, of the l s AaB, AaC, z. AaB is acute.
* Apollonius (260—200 b.c.) studied and probably lectured at Alexandria.
Nicknamed e.
THE TRIANGLE 21
Then, from AABa
AB^ = Aa^ + Ba^ -- 2 Ba . Da,
and from AACa
AC2 :=. tKa? + Ca^ + 2Ca . Da.
But Ca = Ba,
/. AB'^ + AC2 = 2Aa2 + 2Ba2.
Ex. 24. Examine what this theorem becomes in the following cases,
giving a proof in each case :
(i) if A coincides with a point in BC.
(ii) if A coincides with C.
(iii) if A coincides with a point in BC produced.
Ex! 25. The base BC of an isosceles a ABC is produced to D, so that
CD^BC; provethat AD2 = AC2 + 2BC2.
Ex. 26. A side PR of an isosceles a PQR is produced to S so that
RS = PR: prove that QS2 = 2QR2+PR2.
Ex. 27. The base AD of a triangle OAD is trisected in B, C. Prove
that OA2 + 20D2 = 30C2 + 6CD2.
Ex. 28. In the figure of Ex. 27, OA2 + OD2=OB2 + OC2-f 4BC2.
Ex. 29. If Q is a point on BC such that BQ = 7i . QC, then
AB2 + w . AC2 = BQ2 + n . CQ2 + (^^ + 1) aQ2.
(This is a generalized theorem, of which Apollonius' theorem is a par-
ticular case. Also compare Ex. 27.)
Ex. 30. A point moves so that the sum of the squares of its distances
from two fixed points A, B remains constant ; prove that its locus is a circle.
Ex. 31. The sum of the squares on the sides of a parallelogram is equal
to the sum of the squares on the diagonals.
Ex. 32. In any quadrilateral the sum of the squares on the four sides
exceeds the sum of the squares on the diagonals by four times the square on
the straight line joining the mid-points of the diagonals.
22 THE TRIANGLE
Ex. 33. The sum of the squares on the diagonals of a quadrilateral is
equal to twice the sum of the squares on the lines joining the mid-points of
opposite sides.
Ex, 34. In a triangle, three times the sum of the squares on the sides
= four times the sum of the squares on the medians.
Definition. A set of lines which all pass through the same
point are called concurrent.
Definition. A set of points which all lie on the same line
are called coUinear.
Definition. The circumscribing circle of a triangle is often
called the circum-circle ; and its centre the circum-centre.
Theorem 8.
The perpendicular bisectors of the sides of a triangle
are concurrent; and the point of concurrence, s, is the
circumcentre.
Every point on the ± bisector of CA is equidistant from
C and A, and every point on the i. bisector of AB is equidistant
from A and B.
.*. the point where these lines meet is equidistant from A, B,
and C; and, being equidistant from B and C, it is on the ± bisector
of BC.
.*. the ± bisectors of the three sides meet at S, the circum-
centre.
Ex. 35. Through A, B, C draw lines parallel to BC, CA, AB respectively,
forming a triangle A'B'C. By considering the circumcentre of aA'B'C,
prove that the altitudes of a ABC are concurrent.
Ex. 36. Through each vertex of a triangle a pair of lines is drawn
parallel to the lines joining the circumcentre to the other two vertices.
Show that these six lines form an equilateral hexagon, whose opposite
angles are equal.
THE TRIANGLE 23
Definition. The inscribed circle of a triangle is often called
the in-circle ; and its centre the in-centre.
Theorem 9.
The internal bisectors of the angles of a triangle are
concurrent; and the point of concurrence, i, is the in-
centre.
Every point on the internal bisector of Z- B is equidistant from
AB and BC, and every point on the internal bisector of z. C is
equidistant from BC and CA.
.*. the point where these lines meet is equidistant from BC,
CA and AB; and, being equidistant from CA and AB and inside
the triangle, it is on the internal bisector of L A.
.*. the internal bisectors of the three angles meet at I, the
in-centre.
Ex. 37. Prove that r=— .
s
[Use A ABC ^ A I BC + A ICA + A lAB.]
Ex. 38. If a polygon is, such that a circle can be inscribed in it, the
bisectors of the angles are concurrent.
State a corresponding theorem for a polygon about which a circle can be
described.
Ex. 39. Describe a circle to touch a given circle and two of its tangents.
Ex. 40. Prove that any circle whose centre is I cuts off equal chords
from the three sides.
Ex. 41. If Al meets the in-cirele in P, prove that P is the in-centre
of aAYZ. (For notation see p. 16.)
Ex. 42. The internal and external bisectors of z A meet the circumcircle
in K, K'. Prove that KK' is the perpendicular bisector of BC.
Ex. 43. If Al meets the circumcircle in U, SU is perpendicular to BC.
24
THE TRIANGLE
Definition. A circle which touches one side of a triangle, and
the other two sides produced, is called an escribed circle or an
ex-circle. Its centre is called an ex-centre.
A triangle clearly has 3 ex- circles.
Theorem 10.
The internal bisector of l a, and the external bisectors
of ^s B and c are concurrent; and the point of con-
currence is the ex-centre i,.
The proof is left to the reader.
Ex. 44. A, I, h are collinear.
Ex. 45.
are collinear.
Ex. 46. All is L to I2I3.
Ex. 47. If another interior common tangent be drawn to the circles I, h,
and cut BC in K, then IKh is a straight line.
THE TRIANGLE
25
Theorem 11.
A
r = — .
s
B X
C
fig. 15.
AIBC+ AICA+ AIAB =
= A ABC
Now AIBC-i|X.
BC
= lm,
A ICA = Jr6,
A lAB - Ire ;
.*. ^ra + ^rb + ^rc =
-^>
a + b + c
}
.'. rs = A ;
A
28
fig. 17.
(i) AYi + AZi = AC + CYi+AB+BZi
== AC + CXj + AB + BXj (since tangents to a circle
= AC + AB + BC from a point are equal)
= 2s.
But AYi = AZi,
(ii)
(iii)
AY + AZ = AC - CY + AB - BZ
= AC - CX + AB - BX
= AC + AB - BC
= 2s- 2a.
But AY = AZ,
.•. AY = AZ =s-a.
YYi = AYi - AY
= s — {s — a)
= a.
Similarly ZZj = a.
THE TRIANGLE 29
(iv) BXi = BZi-AZi-AB
= S — G.
Also ex - s — c, by proof similar to (ii).
(v) XXi = BC - CX - BXi
— a—2(s — c)
= a-(a + b + c) + 2g
= c-b.
If the figure were drawn with 6 > c, it would be found that
XXi = 6 - c.
Ex. 69. Find the lengths of the segments into which the point of
contact of the in-circle divides the hypotenuse of a right-angled triangle
whose sides are 6 and 8 feet.
Ex. 60. The distance between X and the mid-point of BC is ^ (& ~ c).
Ex. 61. The in-radius of a right-angled triangle is equal to half the
difference between the sum of the sides and the hypotenuse.
Ex. 62. If the diagonals of a quadrilateral ABCD intersect at right
angles at O, the sum of the in-radii of the triangles AOB, BOC, COD, DOA
is equal to the difference between the sum of the diagonals and the semi-
perimeter of the quadrilateral. (Use Ex. 61.)
Ex. 63. Two sides of a triangle of constant perimeter lie along two
fixed lines ; prove that the third side touches a fixed circle.
Definition. The line joining a vertex of a triangle to the
mid-point of the opposite side is called a median.
. Definition. The triangle whose vertices are the mid-points of
the sides is called the medial triangle of the given triangle.
Ex. 64. Prove that two medians trisect one another.
Ex. 65. Hence prove that the three medians are concurrent.
Ex. 66. The circumradius of the medial triangle is ^R.
Lemma 1.
If y, /3 are the mid-points of ab and AC, then yfi is
parallel to BC and equal to JBC.
The proof is left to the reader.
so THE TRIANGLE
Theorem 14.
The medians of a triangle are concurrent ; and each
median is trisected at the point of concurrence, G.
A
fig. 18.
Let the two medians By8, Cy meet at G.
Join Py.
Then, bj^ Lemma 1, y^ is || to BC and =^BC.
Again, As G^y, GBC are similar (?),
.'. G/3 : GB = Gy : GC
= /?y : BC = 1 : 2.
.*. two medians BjS, Cy intersect at a point of trisection
of each.
Let the median Aa cut By8 in G'.
Then it may be proved, as above, that /3g' = JySB, aG' = JaA.
But ^G = lySB,
.*. G' coincides with G,
and aG = JaA.
.*. the three medians are concurrent and each median is
trisected at the point of concurrence, G.
J}^ote. It will be noticed that G, the point of concurrence of
the three medians, is the centroid of the three points A, B, C
(see Chap, in.): accordingly G is called the centroid of the
triangle.
Ex. 67. Prove the centroid property of a triangle by the following
method : let Bj3, Cy meet in G ; produce AG to P so that G P = AG : then
prove that GBPC is a ||«g'"'''» etc.
Ex. 68. The triangles GBC, GCA, GAB are equivalent.
Ex. 69. OnAB, AC points Q, R are taken so that AQ=riAB,AR = i AC.
CQ, BR meet in P, and AP produced meets BC in D ; find the ratio
AP : AD.
Ex. 70. The triangles ABC, a^Sy have the same centroid.
the triangle 31
Theorem 15.
The three altitudes of a triangle are concurrent.
A
:^
B D C
fig. 19.
Draw BE, CF ± to AC, AB ; let them meet in H. Join AH
and produce it to meet BC in D.
We have to prove that AD is ± to BC.
Join FE.
Since z. s AFH, AEH are rt. l s,
A, F, H, E are coney clic ;
.-. L. FAH = L. FEH.
Again, since l s BFC, BEC are rt. l s,
.*. B, F, E, C are concyclic ;
.•. L FEH = L FCB.
But L FAH = L FEH,
.-. L FAH = L FCB,
.*. F, A, C, D are concyclic,
.'. L ADC = L AFC = a rt. L.
Hence AD is ± to BC,
and the three altitudes are concurrent.
Ex. 71. Does the above proof need any modification '\i L^ is right or
obtuse ?
Definition. The point of concurrence, H, of the altitudes of a
triangle is called the orthocentre.
Ex. 72. If H is the orthocentre of a ABC, then A is the orthocentre of
A BCH, B of A CAH, and C of a ABH.
32
THE TRIANGLE
Ex. 73. I is the orthocentre of a Iil2l3'
(Notice that A, I, li are collinear; as also 1^, A, I3.)
Ex.74. AH.HD = BH.HE = CH.HF.
Ex. 75. AS and AH are equally inclined to the bisector of z A.
Ex. 76. Z BHC is the supplement of Z A.
Ex. 77. Show that if two of the opposite angles of a convex quadri-
lateral be right angles, the external diagonal of the complete quadrilateral
formed by the sides is perpendicular to an internal diagonal.
Definition. The triangle whose vertices are the feet of the
altitudes is called the pedal triangle of the given triangle.
Ex. 78. The triangles ABC, HBC, HCA, HAB all have the same pedal
triangle.
Ex. 79. The orthocentre of a triangle is the in-centre of its pedal
triangle.
Ex. 80. The triangle formed by the tangents at A, B, C to the circum-
circle is similar and similarly situated to the pedal triangle.
Theorem 16.
If AH produced meets the circumcircle in x, then
HD = DX.
X
fig. 20.
Since z_ s E and D are rt. l s,
.*. A, E, D, B are concyclic,
.-. Z.DBE=^DAE.
Also L DBX = L DAE, in the same segment.
.*. L DBE = ^ DBX.
Hence As DBH, DBX are congruent,
and HD = DX.
THE TRIANGLE
33
Ex. 81. Draw a figure for Theorem 16, in which z A is obtuse. Does
the proof need any modification for this case ?
Ex. 82. The triangles ABC, AHB, BHC, CHA have equal circum-
circles.
Ex. 83. H is the circumcentre of the triangle formed by the circum-
centres of AHB, BHC, CHA.
Ex.84. BD.DC^AD.HD.
Ex. 85. The base and vertical angle of a triangle are given. Prove that
the locus of the orthocentre is a circle equal to the circurncircle, Find also
the loci of the in-centre and the centroid,
Theorem 17.
AH = 2Sa.
fig. 21,
Let CS meet circurncircle in Q.
Since S and a are the mid-points of CQ and CB respectively,
QB = 2Sa,
and QB is || to Sa and to AH.
Again, as CO is a diameter, l CAQ is a rt. ^ ,
.'. AQis II to HB.
Hence AQBH is a ||««'-'^'".
/. AH = QB = 2Sa.
G. S. M. G. 3
32
THE TRIANGLE
Ex. 73. I is the orthocentre of A lilt's-
(Notice that A, I, Ij are collinear ; as also I2, A, I3.)
Ex.74. AH.HD = BH.HE = CH.HF.
Ex. 76. AS and AH are equally inclined to the bisector of L A.
Ex. 76. z BHC is the supplement of Z A.
Ex. 77. Show that if two of the opposite angles of a convex quadri-
lateral be right angles, the external diagonal of the complete quadrilateral
formed by the sides is perpendicular to an internal diagonal.
Definition. The triangle whose vertices are the feet of the
altitudes is called the pedal triangle of the given triangle.
Ex. 78. The triangles ABC, HBC, HCA, HAB all have the same pedal
triangle.
Ex. 79. The orthocentre of a triangle is the in-centre of its pedal
triangle.
Ex. 80. The triangle formed by the tangents at A, B, C to the circum-
circle is similar and similarly situated to the pedal triangle.
Theorem 16.
If AH produced meets the circumcircle in X, then
HD = DX.
X
fig. 20.
Since jl s E and D are rt. z_ s,
.*. A, E, D, B are concyclic,
.'. Z.DBE=^DAE.
Also z. DBX = ^ DAE, in the same segment.
.'. L DBE = z. DBX.
Hence As DBH, DBX are congruent,
and HD = DX.
THE TRIANGLE
33
Ex. 81. Draw a figure for Theorem 16, in which z A is obtuse. Does
the proof need any modification for this case ?
Ex. 82. The triangles ABC, AHB, BHC, CHA have equal circum-
circles.
Ex. 83. H is the circumcentre of the triangle formed by the circum-
centres of AHB, BHC, CHA.
Ex.84. BD.DC = AD.HD.
Ex. 85. The base and vertical angle of a triangle are given. Prove that
the locus of the orthocentre is a circle equal to the circumcircle, Find also
the loci of the in-centre and the centroid.
Theorem 17.
AH = 2Sa.
fig. 21.
Let CS meet circumcircle in Q.
Since S and a are the mid-points of CQ and CB respectively,
QB = 2Sa,
and QB is || to Sa and to AH.
Again, as C(3l is a diameter, l CAQ is a rt. z. ,
.-. AQ is II to HB.
Hence AQBH is a yogram
.*. AH = QB = 2Sa.
G. S. M. G. 3
84
THE TRIANGLE
Ex. 86. Prove Theorem 17 by using the fact that H is the circumcentre
of the triangle formed by drawing parallels to the sides through the opposite
vertices.
Ex. 87. Let P be the mid-point of AH. Show that aP, SH bisect one
another.
Ex. 88. Show that N, the mid-point of HS, is the centre of the
circle PDa.
Ex. 89. Show that aP is equal to the circumradius of ABC.
Ex. 90. Show that a circle with centre N (the mid-point of HS) and
radius equal to |R will pass through D, E, F, a, j8, 7 and the mid-points of
HA, HB, HC.
Ex. 91. The perpendicular bisectors of Da, E/3, F7 are concurrent.
Ex. 92. Prove that AS, Ha meet on the circumcircle,
Ex. 93. If P, Q, R are the mid-points of HA, HB, HC, then a PQR is
congruent with a a/37.
Ex. 94. SP is bisected by the median Aa.
Ex. 95. The circumradius of a a/37 is JR.
Ex. 96. Prove that A s a^y, D7/3 are congruent.
Ex. 97. Show that a^yD are concyclic. Use Ex. 96 to show that the
circumcircle of the pedal triangle passes through the mid-points of the sides.
Ex. 98. Apply the result of Ex. 97 to the triangle HBC.
Ex. 99. Combining the two preceding exercises, deduce the result of
Ex. 90.
Theoeem 18.
The points H, G, s are collinear ; and HG ^ 2GS.
A
THE TRIANGLE
35
Let Aa cut HS in G'.
Since AH and Sa are ||, As AHG', aSG' are similar,
and since AH = 2Sa,
.'. AG' = 2G'a.
.'. G' is identical with G, the centroid.
Also HG ==2GS.
Ex. lOO. Use fig. 22 to prove, independently, the concurrence of the
three medians.
Ex. lOl. If AS, Ha meet at K, the centroid of A AKH is G.
Theorem 19.
A circle whose centre is the mid-point of SH, and
whose radius is Jr, passes through
D, E, F the feet of the altitudes,
a, p, y the mid-points of the sides,
p, Q, R the mid-points of ha, hb, hc.
Join aP, SH. Let them intersect at N.
(i) H P = iH A = aS, and H P is II to aS,
.-. HPSaisa \\^«^^^\
and the diagonals HS, Pa bisect one another.
.". N is the mid-point of HS and bisects Pa.
3—2
36
D d
fig. 23.
(ii) Since ^ PDa is a rt. zi , Pa is the diameter and N the
centre of 0 PDa.
(iii) AP is equal and || to Sa,
.'. APaS is a \\^sr^^^
:. aPr=SA,
and NP, the radius of 0 PDa = JaP = \Sk = iR.
(iv) It has been shown that the circle whose centre is N,
the mid-point of SH, and whose radius is ^R, passes through the
foot of one altitude, the mid-point of one side, and the mid-point
of HA.
By similar reasoning this circle luay be shown to pass through
the feet of the three altitudes, the mid-points of the three sides,
and the mid-points of HA, HB, HC.
This circle is called the nine-points circle, and its centre N
is called the nine-points centre.
fig. 24.
THE TKIANGLE 37
Ex. 102. The circumcircle of a ABC is the 9-points circle of A li I2 Is-
Ex. 103. The circumcircle bisects each of the 6 lines joining pairs of
the points I, l^, I2, I3.
Ex. 104. If O be equidistant from li, I2, I3, then S is the mid-point
of 01.
Ex. 105. What is the 9-points circle of a BHC ? ::: ^p^Ooi' ^ ^^
Ex. 106. P is any point on the circumcircle of a ABC. PL, PM, PN
are l to BC, CA, AB respectively. Prove that
(i) zPNL = 180° - z PBC. ^
(ii) zPNM=zPAM.
(iii) ZPNL+ zPNM = 180°.
(iv) LNM is a straight line.
Theorem 20.
If from P, a point on the circumcircle, perpendiculars
PL, PM, PN be drawn to the sides of a triangle, then lmn
is a straight line (the Simson* line).
fig. 25.
Join LN, NM.
Since ls PNB, PLB are rt. l s,
.*. L PNL =::180°-Z. PBC.
Again, since l s PNA, PMA are rt. l s,
.-. z.PNM = z.PAM.
But ^PAM-=180°-Z.PAC
= Z. PBC,
.'. z. s PNL, PNM are supplementary,
.". LNM is a straight line.
* Eobert Simson (1687-1768), professor of mathematics at Glasgow ;
author of several works on ancient geometry, and, in particular, of an edition
of Euclid's Elements on which most modern editions are based.
38 THE TRIANGLE
Ex. 107. State and prove a true converse of Th. 20.
Ex. 108. Draw a figure for Th. 20 with P on arc BC ; does the proof
need any modification ?
Ex. 109. What is the Simson Hne of A ? of the point on the circum-
circle diametrically opposite to A ?
Ex. IIO. AD meets the circumcircle in X; the Simson line of X is
parallel to the tangent at A.
Ex. 111. Al meets the circumcircle in U ; the Simson line of U bi-
sects BC.
Ex. 112. If PL meets the circumcircle in U, AU is parallel to the
Simson line.
Ex. 113. The altitude from A is produced to meet the circumcircle in
X, and X is joined to a point P on the circumcircle. PX meets the Simson
line of P in R ; and BC in Q. Prove that R is the mid-point of PQ.
Ex. 114. In Ex. 113 show that HQ is parallel to the Simson line of P.
Ex. 115. From Ex. 114 deduce that the line joining a point on
the circumcircle to the orthocentre is bisected by the Simson line of
the point.
Ex. 116. Prove the last exercise with the following construction : take
image 2? of P in BC ; join ^H, PX, and prove pH parallel to the Simson line
of P.
Ex. 117. Given four straight lines, find a point such that its projections
on the four lines are collinear.
Ex. 118. Given four straight lines, prove that the circumcircle of the
four triangles formed by the lines have a common point. Show that this is
the focus of the parabola that touches the four lines.
THE TRIANGLE 39
Exercises on Chapter IV.
Ex. 110. Given the base, the circumradius, and the difference of the
base angles of a triangle, show how to construct the triangle.
Ex. 120. Two vertices B, C of a triangle are fixed, and the third vertex
A moves in a straight line through B. Prove that the locus of the ortho-
centre is a straight line. What is the locus of the circumcentre ? of the in-
centre ? of the centroid ? of the point where the altitude from A meets the
circumcircle ?
Ex. 121. If a series of trapezia be formed by drawing parallels to the
base of a triangle, the locus of the intersections of the diagonals of these
trapezia is a median of the triangle.
Ex. 122. The base BC of a triangle ABC is divided at P, so that
mBP = nPC ; prove that
mAB2 + n AC2 = (m + n) (AP2+BP. PC).
Ex. 123. The lines joining the circumcentre to the vertices of a triangle
are perpendicular to the sides of the pedal triangle.
Ex. 124. Construct a triangle, given :
(i) two sides and a median (2 cases),
(ii) a side and two medians (2 cases),
(iii) the three medians,
(iv) the base, the difference of the two sides, and the difference of
the base angles,
(v) the base, a base angle, and the sum or difference of the two
other sides,
(vi) the base, the vertical angle, and the sum or difference of the
two other sides,
(vii) the feet of the three perpendiculars,
(viii) an angle, an altitude and the perimeter (2 cases),
(ix) a side, one of the adjacent angles, and the length of the bisector
of this angle,
40 THE TRIANGLE
(x) the sum of two sides, and the angles,
(xi) the perimeter and the angles,
(xii) an angle, the length of its bisector, and one of the altitudes
(2 cases),
(xiii) the angles and an altitude,
(xiv) the base, the sum of two other sides, and the difference of the
base angles.
Ex. 125. Construct a triangle having given the orthocentre, the circum-
centre, and the position (not length) of one of the sides.
Ex. 126. Construct a triangle given the circumcircle, the orthocentre
and one vertex.
Ex. 127. The magnitude of the angle A of a triangle ABC, and the
lengths of the two medians which pass through A and B are known. Con-
struct the triangle.
Ex. 128. The median through A of the triangle AEF makes the same
angle with AB as does Aa with AC.
Ex. 129. If perpendiculars OX, OY, OZ be drawn from any point O to
the sides BC, CA, AB of a triangle,
BX2 + CY2 + AZ2 = CX2 + AY2+BZ2.
State and prove a converse theorem.
Ex. 130. If liX, I2Y, I3Z be drawn perpendicular to BC, CA, AB respec-
tively, these three lines are concurrent.
Ex. 131. Let A I produced meet the circumcircle in K. Prove that
KB = KC = KI.
Draw KK', a diameter of the circumcircle ; and draw lY 1 to AC. Prove
that AS K'KC, AIY are similar.
Hence show that IA.IK=:2Rr; i.e. that the rectangle contained by the
segments of any chord of the circumcircle drawn through the incentre = 2Rr.
Ex. 132. From Ex. 131 deduce that S|2=R2_2Rr.
Ex. 133. Upon a given straight line AB any triangle ABC is described
having a given vertical angle ACB. AD, BE are the perpendiculars from
A, B upon the sides BC, CA meeting them in D and E respectively. Prove
that the circumcentre of the triangle CED is at a constant distance from
DE.
THE TRIANGLE 41
Ex. 134. The triangle ABC has a right angle at C, and AEFB, ACGH
are squares described outside the triangle. Show that if K be taken on AC
(produced if necessary) so that AK is equal to BC, then A is the centroid of
the triangle HEK.
Ex. 135. If four circles be drawn, each one touching three sides of a
given quadrilateral, the centres of the four circles are concyclic.
Ex. 136. The orthocentre of a triangle ABC is H, and the midde point
of BC is D. Show that DH meets the circumcircle at the end of the diameter
through A, and also at the point of intersection of the circumcircle with the
circle on AD as diameter.
Ex. 137. ABC is a triangle, right-angled at A; DEF is a straight line
perpendicular to BC, and cutting BC, CA, AB in E, F, D respectively. BF,
CD meet at P. Find the locus of P.
Ex. 138. Two fixed tangents OP, OQ are drawn to a fixed circle; a
variable tangent meets the fixed tangents in X, Y. Prove (i) that the peri-
meter of the triangle OXY is constant, (ii) that XY subtends a fixed angle at
the centre of the circle.
Ex. 139. Prove that z SAH is the difference between the angles B and
C. Hence construct a triangle, having given the nine-points circle, the
orthocentre, and the difference between two of its angles. Is there any
ambiguity ?
Ex. 140. The lines joining I to the ex-centres are bisected by the cir-
cumcircle.
Ex. 141. The circle BIC cuts AB, AC in E, F; prove that EF touches
the in- circle.
Ex. 142. The triangle formed by the circumcentres of AHB, BHC,
CHA is congruent with ABC.
Ex. 143. Through C, the middle point of the arc ACB of a circle, any
chord CP is drawn, cutting the straight line AB in Q. Show that the locus
of the centre of the circle circumscribing the triangle BQP is a straight
line.
Ex. 144. A circle is escribed to the side BC of a triangle ABC touching
the other sides in F and G. A tangent DE is drawn parallel to BC, meeting
the sides in D, E. DE is found to be three times BC in length. Show that
DE is twice AF.
42 THE TRIANGLE
Ex. 145. Two triangles ABC, DEF are inscribed in the same circle so
that AD, BE, CF meet in a point O; prove that, if O be the in-centre of
one of the triangles, it will be the orthocentre of the other.
Ex. 146. If equilateral triangles be described on the sides of a triangle
(all outside or all inside), the lines joining the vertices of the triangle to the
vertices of the opposite equilateral triangles are equal and concurrent.
Ex. 147. If on the sides of any triangle three equilateral triangles be
constructed, the in-centres of these triangles form another equilateral
triangle.
Ex. 148. Construct a triangle having given the centres of its inscribed
circle and of two of its ex-circles.
Ex. 149. The circumcentre of the triangle BI^C lies on the circumcircle
of ABC.
Ex. ISO. Construct a triangle given the base, vertical angle and in-
radius.
Ex. 151. A pair of common tangents to the nine-points circle and cir-
cumcircle meet at the orthocentre.
Ex. 152. On the sides AB, AC of a triangle ABC any two points N, M
are taken concyclic with B, C. If NC, MB intersect in P, then the bisector
of the angle between AP and the line joining the middle points of BC, AP
makes a constant angle with BC.
Ex. 153. Any line from the orthocentre to the circumference of the
circumcircle is bisected by the nine-points circle.
Ex. 154. If P be any point on the circumcircle and parallels to PA, PB,
PC respectively be drawn through a, j3, y, the mid-points-of the sides, prove
that these parallels intersect in the same point on the nine-points circle.
Ex. 155. If perpendiculars are drawn from the orthocentre of a triangle
ABC on the bisectors of the angle A, show that their feet are collinear with
the middle point of BC.
Ex. 156. If two circles are such that one triangle can be inscribed in
the one and circumscribed to the other, show that an infinite number of
such triangles can be so constructed.
Prove that the locus of the orthocentre of these triangles is a circle.
THE TRIANGLE 43
Ex. 157. The triangle ABC has a right angle at A. AD is the perpen-
dicular from A on BC. O, O' are the centres of the circles inscribed in the
triangles ABD, ACD respectively. Prove that the triangle ODO' is similar to
ABC.
Ex. 158. If D, E, F be the feet of the perpendiculars from a point on
the circumcircle upon the sides, find the position of the point so that DE
may be equal to EF.
Ex. 159. From P, a point on the circumcircle of a triangle ABC, perpen-
diculars PL, PM, PN are drawn to the sides. Prove that the rectangles
PL . MN, PM . NL, PN . LM are proportional to the sides BC, CA, AB.
Ex. 160. The Simson line of a point P rotates at half the rate at
which P rotates about the centre of the circle.
Ex. 161. The Simson lines of opposite ends of a diameter of the
circumcircle are at right angles to one another.
Ex. 162. Find the three points on the circle circumscribing the triangle
ABC such that the pedal lines of the points with respect to the triangle are
perpendicular to the medians of the triangle.
Ex. 163. P, Q, R are three points taken on the sides BC, CA, AB
respectively of a triangle ABC. Show that the circles circumscribing the
triangles AQR, BRP, CPQ meet at a point, which is fixed relatively to the
triangle ABC if the angles of the triangle PQR are given.
If PQR is similar to ABC show that this point is the orthocentre of PQR
and the circumcentre of ABC.
Ex. 164. A straight line AB of constant length has its extremities on
two fixed straight lines OX, OY respectively. Show that the locus of the
orthocentre of the triangle CAB is a circle.
Ex. 166. Find the locus of a point such that its projections upon three
given straight lines are collinear.
Ex. 166. The circumcircle of the triangle formed by any three of the
four common tangents to two circles passes through the middle point of the
line joining their centres.
Ex. 167. If one of the angles of the triangle be half a right angle,
prove that the line joining the orthocentre to the centre of the circumcircle
is bisected by the line joining two of the feet of the perpendiculars from
the angles upon the opposite sides.
44 THE TRIANGLE
Ex. 168. B, C are fixed points, A a variable point on a fixed circle which
passes through B and C. Show that the centres of the four circles which
touch the sides of the triangle ABC are at the extremities of diameters of two
other fixed circles.
Ex. 169. The bisector of the angle BAG meets BC in Y ; X is the point
on BC such that BX = YC, XC=BY ; prove that
AX2-AY2 = (AB-AC)2.
Ex. 170. A straight line PQ is drawn parallel to AB to meet the circum-
circle of the triangle ABC in the points P and Q ; show that the pedal
lines of P and Q. intersect on the perpendicular from C on AB.
Ex. 171. From a point P on the circumcircle of a triangle are drawn
lines meeting the sides in L, M, N, and making with the perpendiculars to
these sides equal angles in the same sense. Show that L, M, N are collinear.
What does this theorem lead to when the equal angles are 90° ?
Ex. 172. If, with a given point P, lines LMN, L'M'N' are drawn as in
the preceding exercise, by taking angles 6, d\ prove that the angle between
LMN and L'M'N' is d-d'.
Ex. 173. Prove that the envelope of all lines LMN (see Ex. 171)
obtained from a fixed point P by varying the angle is a parabola with
focus P and touching the sides of the triangle. What relation does the
Simson line bear to this parabola ?
Ex. 174. Prove that all triangles inscribed in the same circle equi-
angular to each other are equal in all respects.
Ex. 175. The altitude of an equilateral triangle is equal to a side of an
equilateral triangle inscribed in a circle described on one of the sides of the
original triangle as diameter.
Ex. 176. ABC, A'B'C are two triangles equiangular to each other
inscribed in a circle AA'BB'CC. The pairs of sides BC, B'C; CA, C'A';
AB, A'B' intersect in a, b, c respectively.
Prove that the triangle abc is equiangular to the triangle ABC.
Ex. 177. Prove that all triangles described about the same circle equi-
angular to each other are equal in all respects.
THE TRIANGLE 45
Ex. 178. If ABC, A'B'C be two equal triangles described about a circle
in the same sense; and the pairs of sides BC, B'C; CA, C'A'; AB, A'B'
meet in a, b, c respectively ; then a, b, c are equidistant from the centre of
the circte.
Ex. 179. P is a point on the circle circumscribing the triangle ABC.
The pedal line of P cuts AC and BC in M and L. Y is the foot of the
perpendicular from P on the pedal line. Prove that the rectangles PY, PC,
and PL, PM are equal.
CHAPTER V.
THE THEOREMS OF CEYA AND MENELAUS.
Lemma 2.
If two triangles have the same height, their areas are
to one another in the ratio of their bases.
The proof is left to the reader.
Theorem 21.
(The Theorem op Ceva*.)
If the lines joining a point o to the vertices of a
triangle abc meet the opposite sides in x, Y, z, then
BX CY AZ
— . — . — = — 1, the sense of lines being taken into
cx AY BZ °
account.
* The theorem was first published by Giovanni Ceva, an ItaHan, in
1678.
THEOREMS OF CEVA AND MENELAUS
47
fig. 26.
By drawing various figures and placing the point O in the
7 possible different regions, the reader may see that of the ratios
, either 3 or 1 must be negative. The product
BX
AZ
CY
CX' AY' BZ
therefore is negative; and, for the rest, it is sufficient to confine
our attention to the numerical values of the ratios.
BX
CX
AABX
^ACX
AOBX
A OCX
AABX- AOBX
Lemma 2
Similarly
AACX
_ AAOB
" AAOC*
CY A BOC
AY
AZ
BZ A COB
BX CY AZ
CX" AY* BZ
A OCX
A BOA
A.COA
1 (numerically),
= — 1 when sense is taken into account.
48 theorems of ceva and menelaus
Theorem 22.
(Converse of Ceva's Theorem.)
If points X, Y, z are taken on the sides BC, CA, ab of a
BX CY AZ
triangle, such that -^ . — . — =^-1, then are ax, by, cz
ex AY BZ ' '
concurrent.
If AX, BY, CZ are not concurrent, let BY, CZ meet in O, and
let AC (produced if necessary) meet BC in X'.
^, BX' CY AZ
^•'«"cX'-AY-^ = -l <C--)
CX AY BZ
BX' BX
.*. — -J ~ — (sense being taken into account).
CX CX
.*. X' coincides with X,
and AX, BY, CZ are concurrent.
BX' BX
Ex. 180. If ^^^r^, = ^i^^ , where sense is not taken into account, can it
be inferred that X' coincides with X ?
Ex. 181. Using Ceva or its converse (be careful to state which you are
using), prove the concurrence
(i) of the medians of a triangle;
(ii) of the bisectors of its angles ;
(iii) of its altitudes.
Ex. 182. If AZ :ZB = AY:YC, show that the line joining A to the
intersection of BY and CZ is a median.
THEOREMS OF CEVA AND MENELAUS 49
Ex. 183. X, X' are points on BC such that BX = X'C. The pomts
Y, Y'; Z, Z' are similarly related pairs of points on CA, AB. If AX, BY,
CZ are concurrent, so also are AX', BY', CZ'.
Ex. 184. The lines joining the vertices to the points of contact of the
in-circle with the opposite sides are concurrent.
Ex. 185. The lines joining the vertices to the points of contact of the
corresponding ex-circles with the opposite sides are concurrent.
Theorem 23.
(The Theorem op Menelaus"^.)
If a straight line cuts the sides of a triangle abc in
BL CM AN , ^- „ .. , .
L, M, N, then — . — . — == + 1, the sense of lines being
' ' ' CL AM BN ' ^
taken into account.
As in Ceva's theorem, the reader may satisfy himself that
of the ratios — , , — , either 2 or 0 are nearative. The
CL AM' BN ^
product therefore is positive. For the rest of the proof the
sense of lines will be disregarded.
Let the perpendiculars from A, B, C upon LMN be of lengths
a, 13, y.
BL _ 13 CM _ y AN _ a
CL~y' AM~a' BN~J8*
. BL CM AN _ . „
. . — . — . — =1 numerically
CL AM BN "^
= + 1 when sense is taken into account.
* Menelaus of Alexandria, about 98 a.d.
G. s. M. G. 4
Then
50 theorems of ceva and menelaus
Theorem 24.
(The Converse of Menelaus' Theorem.)
If points L, M, N are taken on the sides BC, CA, ab
of a triangle, such that — . — . — =+1, then are L, M, N
CL AM BN
collinear.
The proof is left to the reader.
Ex. 186. Prove theorem 24.
Ex. 187. Use the above theorems to prove the theorem of the Simson
line (Th. 20).
[Let z PAB = ^, then AN=APcos^, etc.]
Ex. 188. If points Q, R are taken on AB, AC so that AQ = 2QB,
AR = iRC, and QR produced meets BC in P, find PB : PC.
Ex. 189. The bisectors of z s B and C meet the opposite sides in Q, R,
and QR meets BC in P; prove that AP is the exterior bisector of z A.
Ex. 190. a, /8, 7 are the mid-points of the sides ; Aa meets J87 in P ;
CP meets AB in Q. Show that AQ = i AB.
Exercises on Chapter Y.
Ex. 191. A straight line cuts the sides BC, CA, AB of a triangle in
L, M, N respectively. The join of A to the intersection of BM, CN meets BC
in P. Show that BC is divided in the same ratio at L and P.
Ex. 192. The sides BC, CA, AB of a triangle ABC are divided in-
ternally by points A', B', C so that BA' : A'CzrCB' : B'Arr AC : C'B. Also
B'C produced cuts BC externally in A". Prove that
BA":CA" = CA'2:A'B2.
Ex. 193. Points P, P' are taken on BC such that PB = CP', and CB,
AB, AC are bisected in O, K, L respectively. Prove that the intersections of
OL with AP and of KP with LP' are collinear with B.
Ex. 194. X is any point on llj ; BX, CX meet AC, AB in Q, R ; QR
meets BC in U, Show that UI2I3 is a straight line.
THEOREMS OF CEVA AND MENELAUS 51
Ex. 195. The lines EF, FD, DE, which join the points of contact
D, E, F of the inscribed circle of a triangle with the sides, cut the opposite
sides in X, Y, Z. Prove that X, Y, Z are coUinear.
Ex. 196. A transversal through P, on BC produced, cuts off equal
lengths BR, CQ from the sides AB, AC of a triangle. Show that
PQ:PR = AB:AC.
Ex. 197. If AD, BE, CF are concurrent straight lines meeting the
sides of the triangle ABC in D, E, F respectively, and the circle DEF cuts
the sides again in D', E', F', prove that AD', BE', CF' are concurrent.
Ex. 198. llirough a point F on the diagonal BD of a square A BCD
lines are drawn parallel to the sides to meet AB in G, BC in E, CD in K,
and DA in H. Prove that BH, CF, and DG are concurrent.
Ex. 199. ABC is a triangle right-angled at C ; P is any point on AB.
Perpendiculars are let fall from P on CA and CB. The line joining the feet
of these perpendiculars meets AB in Q. Prove that 2P0. PQ=PA. PB
where O is the mid-point of AB.
Ex. 200. DEF is the pedal triangle of ABC ; O lies on AD ; OE, OF
meet DF, DE in Y, Z. Show that FE, YZ, BC are concurrent.
Ex. 201. S is a point on the side QR of a triangle PQR. The lines
joining S to the mid-points of PQ, PR meet PR, PQ at T, U respectively.
TU meets QR at V. Prove that QV : RV= SQ'-^ : RS^.
Ex. 202. If the in-circle touch AB in Z, and the circle escribed to BC
touch AC in Yi, then ZYi is divided by BC in the ratio AC : AB.
Ex. 203. A line drawn through the vertex A of a square ABCD meets
the sides BC, CD in E and F; DE and BF meet in G ; CG meets AD
in H. Prove that DF^DH.
Ex. 204. The sides A B, CD of a quadrilateral ACDB are parallel; CA,
DB meet in E, CB, AD meet in H, and CB, AD meet FEG, a parallel to AB,
in G and F respectively. Show that AG, BF, and EH are concurrent.
4—2
52 THEOREMS OF CEVA AND MENELAUS
Ex. 205. The line CF cuts the side AB of a triangle ABC in a point F
such that AF : FB=:w : 1; and lines are drawn through A and B parallel to
the opposite sides. Show that the ratios of the area of the triangle formed
by these lines and CF to the area of the triangle ABC is (1 - n)^ : w.
Ex. 206. D, E, F are points on the sides of a triangle ABC, and AD,
BE, CF meet in O. Prove that
OP OE OF
AD"^BE'^CF""
CHAPTER VI.
HARMONIC SECTION.
[Throughout this chapter, the sense of lines will be taken into account.li
Definition. If a straight line AB is divided at two points
AC /AD
C, D so that — / — = -1, it is said to be divided har-
CB/ DB '
monically; A, C, B, D are said to form a harmonic range;
and C and D are called harmonic conjugates with respect
to A and B.
Note that the above definition is the same as the following. If a straight
line is divided internally and externally in the same ratio, it is said to be
divided harmonically.
AC
Ex. 207. Take a line AB 6 cm. long; divide it at C so that =^=2;
Cb
find the point D such that C, D divide AB harmonically.
Ex. 208. Kepeat Ex. 207 with (i) ^ = 1, (ii) ^= -2, (iii) ^= -|.
Ex. 209. If AB is divided harmonically at C, D, then CD is
divided harmonically at A, B.
Ex. 210. Draw a scalene triangle ABO; draw the internal and external
bisectors of the angle at O and let them cut the base in C and D. Calculate
AC /AD
(from actual measurements) ^^ / --. Is A, C, B, D a harmonic range?
Ex. 211. Prove that the internal and external bisectors of an
angle of a triangle divide the opposite side of the triangle har-
monically.
54 HARMONIC SECTION
Definition. If A, C, B, D be any four points in a straight
line, — /— ;: is called their cross-ratio and is written {AB, CD|.
CB/ DB I J )
[The cross-ratio {AB, CD} is the ratio of the ratios in which C and D
divide AB.]
We see that, if {AB, CD} = — 1, A, C, B, D is a harmonic
range.
Theorem 25.
If {AB, CD} - - 1, then -i- + i- = A
^ ' ^ ' AC AD AB
• — 1 1 1
A C B
D
fig. 30.
Let AB = £C, AC == y, AD = 2;.
If {AB, CD} = -1,
^, AC /AD
then ^rzh:-=-\.
CB/ DB
• y 1 ^ - \
" x-yj x-z
. y ^
x — y X — z
:. yx-yz = -xz + yz.
:. xz + yx=2yz.
.1 12
" y z~ x'
1 1 2
I.e. 1 =: — .
AC AD AB
.*. AC, AB, AD are in harmonic progression; hence the name
'harmonic range.*
Ex. 212. Prove that the same property is true for the distances measured
from any one of the four points.
HARMONIC SECTION 55
Theorem 26.
If A B is divided harmonically at c, D, and if o is the
midpoint of ab, then oc . OD =: OB^
C 8
;. 31.
Let OB = b, OC = c, OD = c? ; then AO = h.
If {AB, CD} = -1,
then^/^=.-l.
. CB/ DB
b + G /b + d
b-cj b-d
. 6 + c b + d
' ' b — G b — d'
.'. b'^ + bc — bd-cd^-b^-bd + bc + cd.
:. 2b''=2cd.
:. ¥ = cd.
i.e. OB^r. OC.OD.
Ex. 213. Prove the converse of the above proposition, namely, that if
O is the mid-point of AB and OC . 0D = 0B2, then {AB, CD} = - 1.
Ex. 214. If {AB, CD} = - 1 and P is the mid-point of CD, then
PA . PB = PC2.
Ex. 215. If AB is divided harmonically at C, D and if O is the mid-
point of AB and P of CD, prove that OB2+ PC2 = OP2.
Ex. 216. If {AB, CD} = - 1, what is the position of D when C coincides
with (i) A, (ii) the mid-point of AB, (iii) B, (iv) the point at infinity.
Ex. 217. Prove that if ACBD be a harmonic range and if O be the
middle point of CD, then AC is to CB as AO to CO.
66 HARMONIC SECTION
Ex. 218. P, Q divide a diameter of a circle harmonically; P', Q' divide
another diameter harmonically ; prove that P, P', Q, Q' are concyclic.
Ex. 219. If X, Y, Z are the points at which the in-circle of a triangle
ABC touches the sides, and if YZ produced cuts the opposite side in X',
then X and X' divide that side harmonically.
[Use Menelaus' Theorem.]
Ex. 220. Prove the same theorem for the points of contact of one
of the ex- circles.
Ex. 221. On a straight line take four points A, C, B, D such that
AC = l-6 in., CB = 0-8 in., BD = 2-4 in. What is the value of {AB, CD}?
Take any point O outside the line. Draw a straight line parallel to CD
cutting OA, OB, OC at P, Q, R. Find experimentally the value of the
ratio PR/RQ.
Again draw parallels to OA, OB or OC in turn, and try to discover
a law.
Ex. 222. {AB, CD}=-1; O is any point outside the line ACBD ;
through C draw PCQ parallel to OD cutting OA, OB at P, Q. Prove
PC=:CQ
[By means of similar triangles express PC/OD in terms of segments of
the line ACBD, and then express CQ/OD in the same way;]
Ex. 223. Prove the converse of Ex. 222, namely, that if PC = CQ
thenjAB, CD}=-1.
Ex. 224. Draw ACBD as in Ex. 221; take any point O outside the line
and join OA, 00, OB, OD; draw a line cutting these lines at A', C, B', D';
measure and calculate {A'B', CD'}. Repeat the experiment for another
position of A'C'B'D'.
Ex. 225. If a point O be joined to the points of a harmonic range
ACBD and these lines be cut by a straight line in A', C, B', D' ; prove
that {A'C'B'D'} is harmonic.
[Through C and C draw parallels to OD, and use Exs. 222, 223.]
Definition. A system of lines through a point is called a
pencil. The point is called the vertex of the pencil.
Definitio7b. Any straight line drawn across a system of lines
is called a transversal.
HARMONIC SECTION
67
Theorem 27.
If a transversal cuts the four lines of a pencil at
A, c, B, D, and if AC BD is a harmonic range, then any
other transversal will also be divided harmonically.
5r /
\,
A/ C/
B
\D
Q
\
/
B"^ —
— — —_____^
<?
fig. 32.
Let O be the vertex of the pencil.
Let a straight line cut the rays of the pencil at A', C', B', D'.
Through C and C' draw straight lines n to OD, cutting OA
at P, P' and OB at Q, Q'.
( ': PC is II to OD.
.*. As APC, AOD are similar.
AC PC
•• AD ~0D'
also As BQC, BOD are similar.
CB QC
* * DB ~ O 5 '
Since {AB, CD} = -1,
AC /AD
* ' CB/ 5 B
AC_ CB
" AD~~DB
PC _ QC
db~~OD
.-. PC = CQ.
[AB, CD}= -1
.*. PC = CQ
-1.
58
HAKMONIC SECTION
O
fig. 32.
P'C = C'Q'
fA'B', C'D'
Now
P'C OC C'Q'
PC OC CQ
.*
. P'C' :- C'Q'.
^nd from
similar triangles as before
A'C P'C
A'D' ~ OD"
, C'B' Q'C
and — r— , = 7 .
D'B' OD'
^-li
A'C' C'B'
A'D' ~ D'B"
. A'C /A'D'
C'B'/ D'B' ~
• 1
A'B,' C'D'} is a harmonic range.
Definitio7i. If a pencil of four lines divides one transversal
(and therefore every transversal*) harmonically, the pencil is
called a harmonic pencil.
0{AB, CD} = - 1 denotes that the pencil OA, OB, OC, CD is
harmonic; OC and CD are called harmonic conjugates with
respect to OA and OB.
* This follows from the proposition just proved.
HARMONIC SECTION 59
Note on Theorem 27.
From Theorem 26 it is easy to show that if C and D are harmonic
conjugates with respect to AB, and if D is at infinity, then C is the mid-
point of AB. [See also p. 8, § 3 (iii).]
In the course of proving Theorem 27 we saw that a transversal PQ
parallel to OD is bisected by OC ; it should be noticed that this is a
particular case of the theorem; for, since PQ is parallel to OD it cuts
OD at infinity; therefore C and the point at infinity are harmonic con-
jugates with respect to PQ; therefore C is the mid-point of PQ.
Theorem 28.
The internal and external bisectors of an angle are
harmonic conjugates with respect to the arms of the
angle.
The proof is left to the reader.
Theorem 29.
If {AB, CD} = - 1 and o is a point outside the line such
that z. COD is a right angle, then oc, OD are the bisectors
of Z.AOB.
The proof is left to the reader.
Ex. 226. "What line is the harmonic conjugate of the median of a
triangle with respect to the two sides through the vertex from which the
median is drawn ?
Ex. 227. If a, ^, 7 are the mid-points of the sides of a triangle ABC,
prove that a {7/8, AC} := - 1.
Ex. 228. If D, E, F are the feet of the altitudes of a triangle ABC,
prove that D{EF, AB}=: -1.
Ex. 229. If X, Y, Z are the points of contact of the in-circle and the
sides of the triangle ABC, prove that XJYZ, AC}= - 1.
60 HAEMONIC SECTION
Ex. 230. Lines are drawn parallel to the sides of a parallelogram
through the intersection of its diagonals ; prove that these lines and the
diagonals form a harmonic pencil.
Ex. 231. If A, B, C, D and A', B', C, D' are two harmonic ranges and
if AA', BB', CC all pass through a point O, then O, D, D' are coUinear.
Exercises on Chapter YI.
Ex. 232. A, B, C, D, O, P are points on a circle and O {AB, CD} = - 1 ;
prove that P{AB, CD} = - 1.
Ex. 233. The bisector of the angle A of a triangle ABC meets BC in X ;
prove that AX is divided harmonically by the perpendiculars drawn to it
from B and C.
Ex. 234. The pencil formed by joining the four angular points of a
square to any point on the circumscribing circle of the square is harmonic.
Ex. 235. A chord AB and a diameter CD of a circle cut at right angles.
If P be any other point on the circle, P (AB, CD) is a harmonic pencil.
Ex. 236. a, ^, y are the mid-points of the sides of a triangle ABC,
Aa and ^y intersect at X, a line is drawn through X cutting 07, a/3, BC at
Y, Z, W respectively; prove that Y, X, Z, W form a harmonic range.
Ex. 237. Three lines pass through a point ; through a given point on
one of the lines draw a line that shall be divided into two equal parts by
the other two.
Ex. 238. Find a point P in a given straight line so that the lines
joining P to three given points in a plane containing the given line may cut
off on any line parallel to the given line and lying in the same plane two
equal segments.
Ex. 239. If X, Y, Z are points on the sides BC, CA, AB of a
triangle such that AX, BY, CZ are concurrent, and if YZ meets BC
in X' ; then is {BC, XX'} a harmonic range.
Ex. 240. TP, TQ are two tangents to a circle; PR is a diameter of the
circle and QN is drawn perpendicular to PR. Prove that Q{TPNR} is a
harmonic pencil.
Ex. 241. In a triangle ABC the line AD is drawn bisecting the angle A
and meeting BC in D. Find a point P in BC produced either way, such
that the square on PD may be equal to the rectangle PB . PC.
HARMONIC SECTION 61
Ex. 242. P is a point on the same straight line as the harmonic range
ABCD ; prove that
PA PB Pp
AC~BC^DC*
Ex. 243. A, B, C, D are four points in a straight line ; find two points
in the line which are harmonic conjugates with respect to A, B and also
with respect to C, D.
Ex. 244. ABC is a triangle ; through D, the mid-point of BC, a
straight line PDQR is drawn cutting AB, AC in P, Q respectively. AR
is drawn parallel to BC, and cuts BQ at S. Prove that AR = RS.
Ex. 245. PAQB is a harmonic range, and a circle is drawn with AB as
diameter. A tangent from P meets the tangent at B in S, and touches the
circle in T. Prove that SA bisects TQ.
Ex. 246. Through* one angle O of a parallelogram OEAF a line is
drawn meeting AE and AF, both produced, in B and C respectively.
Prove that the area A EOF is a harmonic mean between the areas BOA
and COA.
Ex. 247. TP, TQ are two tangents to a circle ; PR is a diameter of the
circle and QN is drawn perpendicular to PR. Prove that QN is bisected
byTR.
Ex. 248. If X is any point in AD an altitude of a triangle ABC, and
BX, CX produced cut the opposite sides of the triangle in Y and Z, then
Z YDZ is bisected by DA.
Ex. 249. Prove that the lines joining any point on a circle to the ends
of a fixed chord cut the diameter perpendicular to the chord in two points
which divide the diameter harmonically.
Ex. 250. If A', B', C lie on the sides BC, CA, AB of a triangle and
AA', BB', CC be concurrent; and if A" be the harmonic conjugate of A'
with respect to B, C while B", C" are similarly determined on the other
sides ; then A", B", C" are collinear.
Ex. 251. The lines VA', VB', VC bisect the internal angles formed by
the lines joining any point V to the angular points of the triangle ABC ;
and A' lies on BC, B' on CA, C on AB. Also A", B", C" are harmonic con-
jugates of A', B', C with respect to B and C, C and A, A and B. Prove that
A", B", C" are collinear.
Ex. 252. The inscribed circle of a triangle ABC touches the sides BC,
CA, AB in A', B', C respectively. Show that the points in which the
circles B'C'B and B'C'C meet BC again are equidistant from A'.
CHAPTER VII.
POLE AND POLAR.
Definition. The line joining the points at which two tangents
touch a circle is called their chord of contact.
Ex. 253. If tangents are drawn to a circle from an external point, the
line joining this point to the centre of the circle bisects the chord of contact
at right angles.
Ex. 254. What is the chord of contact of a point on the circumference?
Provisional Definition. If P and Q are the points of contact
of the tangents to a circle from a point T, the straight line
through P and Q is called the polar of T, and T is called the
pole of PQ, with respect to the circle.
This definition of the polar of a point is meaningless when the point is
inside the circle. It will therefore be necessary to find a new definition. But
before doing so we must prove the following theorem.
POLE AND POLAR
63
Theorem 30.
If the line joining a point T to the centre c of a circle
cuts the chord of contact of T in N and the circle in A,
then CN . CT == ca2.
fig. 33.
Let P and Q be the points of contact of the tangents from T.
Then, in the As CPN, CTP, z. C is common,
and z. s CNP, CPT are equal (being rt. l s).
.*. the As are similar.
CNCP
CP ~ cf*
.-. CN.CT = CP2-^CA2.
Definition. If T and N are two points on a line, drawn from
C the centre of a circle, such that CN . CT is equal to the square
on a radius of the circle, and if through N a line XY is drawn at
right angles to CN, XY is called the polar of T and T is called
the pole of XY with respect to the circle.
fig. 34.
64
POLE AND POLAR
Ex. 255. Prove that this definition agrees with the 'provisional defini-
tion' when T is outside the circle.
Ex. 256. What is the position of the polar in the following cases: (i) T
on the circle, (ii) T coinciding with C, (iii) T at infinity ?
Ex. 257. A and B are two concentric circles; what is the envelope of
the polar with respect to A of a point which moves round B ?
Ex. 258. What are the polars of the vertices of a triangle with respect
to (i) its incircle, (ii) its circumcircle ?
Ex. 259. ABC is a triangle. A circle is described with A as centre and
radius AX such that AX^^ AB . AF where F is the altitude from C. What
lines are the polars of B and C? and what point is the pole of BC?
Ex. 260. If firoxn a fixed point T any line is drawn cutting a
circle in R and S, prove that tlie tangents at R and S intersect on
a fixed line (viz. the polar of T).
[Let X be the point of intersection of the tangents; draw XN j. to
the line joining T to the centre C ; let CX cut RS in K. Prove that
CN .CT = CK.CX = CS2.]
Ex. 261. Prove that any point T and the polar of T with respect to a
circle divide the diameter through T harmonically.
Theorem 31.
If a straight line is drawn through any point to cut a
circle, the line is divided harmonically by the circle, the
point and the polar of the point with respect to the
circle.
fig. 36.
fig. 37.
POLE AND POLAR 65
Let T be the point, TRHS the line cutting the circle at R, S
and the polar of T at H ; let C be the centre of the circle, and
let CT cut the circle at A and the polar of T at N.
Draw CK J- to RS ; then K is the mid-point of RS.
[If we can prove KH . KT = KR-, then {RS, HT} = - L]
Sense being taken into account, we see that
KH. KT = KT(KT-HT)
^-KT^-KT. HT.
Now in both figures C, K, H, N are concyclic, because the L s
at K and N are right angles.
.*. KT . HT = CT . NT.
.*. KH .KT = KT2-CT.NT
= KT2-CT(CT-CN)
= KT2-CT2 + CN.CT
= -CK2 + CA2 (CN .CT=zCA2 by def.)
= CR2-CK^
= KR2.
.'. {RS, HT} is harmonic.
Ex. 262. If H be the harmonic conjugate of a fixed point T
with regard to the points in which a line through T cuts a fixed
circle, the locus of H is a straight line.
[Use reductio ad absurdum.]
Ex. 263. If C be the centre of a circle, and the polar of a point T cut
TC in N, and any straight h'ne through T cut the circle in R and S, then
the polar bisects the angle RNS.
Ex. 264. If a straight line TRS cut a circle in R and S and cut the
polar of T in H, and if K be the mid-point of RS, then TR . TS = TH . TK.
Ex. 265. The polar of a point O with regard to a circle meets it in
A, B ; any chord through O meets the circle in C, D. Prove that A, B, C, D
subtend a harmonic pencil at any point of the circle,
[Consider the pencil subtended at A.]
G. s. M. G. 5
66 POLE AND POLAR
Ex. 266. A chord PQ of a circle moves so that the angle it subtends at
a fixed point O inside the circle is bisected externally by the diameter through
O ; prove that PQ passes through a fixed point.
Is this theorem true when O is the centre of the circle ?
Ex. 267. State and prove a theorem corresponding to Ex. 344 for the
case in which the diameter bisects the angle internally, O being still inside
the circle.
Ex. 268. A chord PQ of a circle moves so that the angle it subtends at
a fixed point O outside the circle is bisected by the diameter which passes
through O ; prove that PQ either passes through a fixed point or is parallel
to a fixed direction.
Ex. 269. Prove that, if the points A, B, C, D all lie on a circle, the
polar of the point of intersection of AC, BD passes through the point of
intersection of AB, CD.
[Let AB, CD intersect at X, and AC, BD at Y; find Z the harmonic
conjugate of Y with respect to B, D; let CA meet ZX in T; prove XT the
polar of Y.]
Theorem 32.
If the polar of a point P with respect to a circle passes
through a point Q, then the polar of Q passes through P.
fig. 38.
Let C be the centre of the circle, and QN the polar of P.
From P draw PM x to CQ,
POLE AND POLAR 67
Then Q, M, N, P are coney clic.
.-. CM.CQ = CN.CP
= CAl ( '.• QN is the polar of P)
.*. PM is the polar of Q.
.'. the polar of Q passes through P.
Ex. 270. Sketch a figure for Theorem 32 with both P and Q outside
the circle.
Ex. 271. Prove this theorem by the harmonic property of pole and
polar for the particular case in which PQ cuts the circle.
Ex. 272. If a point moves on a straight line its polar with respect
to a circle passes through a fixed point.
Ex. 273. If a straight line moves so that it always passes
through a fixed point, its pole with respect to a circle moves on a
straight line.
Ex. 274. The line joining any two points A and B is the polar
of the point of intersection of the polar s of A and B.
Ex. 275. AB, AC touch a circle at B, C. If the tangent at any other
point P cuts BC produced at Q, prove that Q is the pole of AP.
Ex. 276. AB, AC touch a circle at B, C ; the tangent at another point
P on the circle cuts BC at Q. Prove that A{BC, PQ} = - 1.
5-2
68
POLE AND POLAR
Theorem 33.
Two tangents are drawn to a circle from a point A on
the polar of a point B ; a harmonic pencil is formed by
the two tangents from A, the polar of B and the line ab.
fig. 39.
Let AP, AQ be the tangents from A.
Since the polar of B passes through A, ,'
(i.e. PQ) passes through B.
Let the polar of B cut PQ at C.
Then P, C, Q, B is a harmonic range.
.'. the pencil AP, AC, AQ, AB is harmonic.
Ex. 277. Prove Ex. 219 by means of Theorem 33.
the polar of A
[Th. 31.]
Ex. 278. From any point P on a fixed straight line XY tangents
I^Z, PW are drawn to a circle ; prove that, if PT is such that the pencil
P{ZW, YT} is harmonic, PT passes through a fixed point.
[Prove that the intersection of ZW, PT is the pole of XY.]
Ex. 279. Prove that, if the lines PX, PY, QX, QY all touch a circle,
then XY passes through the pole of PQ.
[Draw PZ to cut XY at Z, such that P{XY, ZQ}= -1; and consider
the pencil Q{XY, ZP}.]
POLE AND POLAR 69
An interesting case of pole and polar is that in which the
circle has an infinite radius.
T A
fig. 40.
Let AB be a diameter of a circle, T any point on it and let
the polar of T cut AB at N ; then TN is divided harmonically at
A and B.
Now suppose that A and T remain fixed and that B moves
along the line TA further and further from B ; in the limit when
B has moved to an infinite distance, TA = AN (since {TN, AB} = — 1);
and the circle becomes the line at infinity together with the line
through A at right angles to TN.
Thus the polar of a point T with respect to a line (regarded
as part of a circle of infinite radius) is a parallel line whose
distance from T is double the distance of the given line from T.
Ex. 280. Into what do the following properties degenerate in the case
in which the circle has an infinite radius: (i) Theorem 31, (ii) Ex. 262,
(iii) Ex. 272, (iv) Ex. 273 ?
Exercises on Chapter YII.
Ex, 281. Through a point A within a circle are drawn two chords
PP', QQ' ; show that PQ, P'Q' subtend equal angles at B, the foot of
the perpendicular from A to the polar of A with respect to the circle.
Ex. 282. TP, TQ are two tangents to a circle ; prove that the tangent
to the circle from any point on PQ produced is divided harmonically by TP
and TQ.
Ex. 283. The tangents at two points P and Q of a circle intersect at
T ; HTK is drawn parallel to the tangent at a point R, and meets PR and
QR in H and K respectively ; prove that HK is bisected in T.
70 POLE AND POLAR
Ex. 284. From a point O a line is drawn cutting a circle in P and R
and the polar of O in Q ; if N is the mid-point of PR and if the polar of O
meets the circle in T and T', show that the circles TQN, T'QN touch OT,
OT' respectively.
Ex. 285. A fixed point A is joined to any point P on a circle, AQ is
drawn to cut the tangent at P in Q so that L PAQ=: z APQ, AQ is pro-
duced to R and QR = AQ ; prove that R lies on the polar of A.
Ex. 286. From a point O are drawn two straight lines, OT to touch a
given circle at T and OC to pass through its centre C, and TN is drawn to
cut OC at right angles in N. Show that the circle which touches 00 at O
and passes through T cuts the given circle at a point S such that the straight
line TS produced bisects NO.
Ex. 287. AOB, COD are chords of a circle intersecting in O. The
tangents at A and D meet in P, and the tangents at B and 0 meet in Q.
Show that P, O, Q are collinear.
Ex. 288. The product of the perpendiculars on any two tangents to a
circle from any point on its circumference is equal to the square on the
perpendicular from the point to the chord of contact.
Ex. 289. I is the centre of the incircle of a triangle ABC; lines through I
perpendicular to I A, IB, 10 meet the tangent at P to the incircle in D, E, F
respectively. Find the positions of the poles of AD, BE, OF with respect
to the incircle: and hence (or otherwise) prove that these three lines are
concurrent.
Ex. 290. The distances of two points from the centre of a circle are in
the same ratio as their distances each from the polar of the other with
respect to the circle (Salmon's theorem).
Ex. 291. The harmonic mean of the perpendiculars from any point O
within a circle to the tangents drawn from any point on the polar of O is
constant.
CHAPTER VIII.
SIMILITUDE.
1. In elementary geometry* we have seen that, if a point
O is joined to each vertex of a given polygon, and if each of the
joins is divided in the same ratio, these points of division are
the vertices of a similar polygon.
Extending this principle, we see that, if a 'point O is joined to
a 'point P, and OP is divided in a fixed ratio at Q, as P describes
a given figure {consisting of any number of lines and curves), the
point Ql will describe a similar figure.
Ex. 292. Draw a circle of radius 4 cm. ; mark a point O, 10 cm. from
its centre ; if P is any point on the circle plot the locus of the mid-point
of OP.
Ex. 293. Prove that the locus is a circle in Ex, 292.
Ex. 294. P is a variable point on a fixed circle whose centre is O; a
point Q is taken on the tangent at P, such that angle POQ is constant;
what is the locus of Q?
Ex. 295. Draw a triangle ABC having BC = 8 cm., CA = 6 cm.,
AB = 7 cm. ; mark a point P 4 cm. from B and 6 cm. from C. The triangle
is now rotated about P through a right angle, to the position abc ; explain
how you determine the points a, b, c and find what angle ac makes with AC.
2. If a figure is rotated about a point O through any angle a,
the angle through which any line in the figure has been rotated
(i.e. the angle between the new position and the old) is a.
* G-odfrey and Siddons' Elementary Geometry, iv. 9.
72 SIMILITUDE
3. Again suppose O a fixed point and P any point on a given
figure^ and Q a point such that OQ : OP = k: 1, and L POQ = a
{k and a being constants); as P describes any figure, Q will
describe a similar figure.
For suppose Q' a point in OP such that OQ! : OP = k : I, Ql'
will describe a figure similar to the " P" figure; and if we now
rotate the " Q'" figure about O through an angle a it will
coin-cide with the "Q" figure.
Ex. 296. P is a variable point on a fixed circle, O any point inside it ;
PQ is drawn at right angles to OP and OQ makes a fixed angle (always
taken in the same sense) with OP. What is the locus of Q?
4. If ABC, DEF are two similar triangles with their corre-
sponding sides parallel, then AD, BE, OF will be concurrent.
For if AD and BE cut at O, OA : OD == AB : DE; and if AD and
OF cut at O', O'A : O'D = AC : DF = AB : DE = OA : OD ; .-. O and O'
coincide.
Extending this we see that, if ABC D ... A'B'C'D' ... are two
similar rectilinear figures with their corresponding sides parallel,
AA', BB', CC', DD', ... are concurre7it ; or going a step further we
see that the same is true even when the figures consist of curves as
well as straight lines.
When two similar figures are so placed that the join of each
pair of points in the one figure is parallel to the join of the cor-
responding pair of points in the other figure, the two figures are
said to be similarly situated and the point of concurrence of
SIMILITUDE 73
the lines joining corresponding points is called the centre of
similitude.
In the case of triangles we have seen that AD is divided
(externally or internally) in the ratio of AB : DE.
So in the general case the centre of similitude divides the joins
oj corresponding points in the ratio of the linear dimensions of the
two figures.
Ex. 297. Draw a careful figure of two similar and similarly situated
circles; indicate several corresponding points and draw the tangents at a
pair of such points.
Ex. 298. Draw (or plot) an accurate parabola. Draw a similar and
similarly situated curve, (i) when the centre of similitude is on the axis,
(ii) when it is not on the axis.
Ex. 299. If any line through the centre of similitude of two
curves cuts them at corresponding points P and P', the tangents at
P and P' are parallel.
Or in other words, any line through the centre of similitude of
two curves cuts them at the same angle at corresponding points.
[Use the method of limits.]
Ex. 300. If O is a centre of similitude of two curves, a tangent
firom O to either of the curves touches the other curve, and the points
of contact are corresponding points.
fig. 43. fig. 44.
5. In the case of two circles there are two centres of similitude^
viz. the points which divide the line joining the centres externally
and internally in the ratio of the radii.
In fig. 44 the constant ratio is negative.
74 SIMILITUDE
Ex. 301. If a circle A touches two circles B, C at points P, Q,
prove that PQ passes through a centre of similitude of the circles
B, C.
Note that there are two cases.
Ex. 302. Prove that the common tangents to two circles pass
through one of their centres of similitude.
Ex. 303. What is the centre of similitude of a line and a circle ?
Have they two centres of similitude?
Ex. 304. Have two parallel lines a centre of similitude?
Ex. 305. Have two intersecting lines a centre of similitude?
Exercises on Chapter VIII.
Ex. 306. A triangle ABC is given in specie (i.e. its angles are given)
and the point A is fixed ; prove that B and C describe similar loci.
Ex. 307. Find the centres of similitude of the circumcircle and nine-
points circle of a triangle.
OX
Ex. 308. O is a fixed point, XOY a constant angle, p— r a constant ratio.
Find the locus of Y when that of X is (i) a straight line, (ii) a circle.
Ex. 309. Prove that the line joining the vertex of a triangle to that
point of the inscribed circle which is furthest from the base passes through
the point of contact of the escribed circle with the base.
Ex. 310. A triangle ABC is inscribed in a given circle, and its vertex
A is fixed. Show that the locus of a point P on BC, such that the ratio of
AD 2 to BD . DC is given, is a circle touching the given circle at A.
Ex. 311. C is a moving point on a circle of which O is centre and
AB is a fixed diameter; BC is produced to D so that BC = CD. Find the
locus of the intersection of AC and CD.
Ex. 312. In a quadrilateral ABCD, the points A and B are fixed, and
the lengths BC, CA and CD are given. Find the locus of (1) the mid-point
of BD, (2) the mid-point of the line joining the mid-points of the diagonals.
Ex. 313. Through a point O draw a line cutting a circle in P, Q, such
that the chord PQ is i of OQ.
SIMILITUDE 75
Ex. 314. A is a moving point on a fixed diameter BD (produced) of a
circle; AC is a tangent from A; P is the projection of the centre on the
bisector of the angle OAC. Find the locus of P.
Ex. 315. Inscribe in an equilateral triangle another equilateral triangle
having each side equal to a given straight line.
Ex. 316. Describe a triangle of given species (given angle) so that one
angular point may be at a given point and the others on given straight
lines.
Ex. 317. O is a fixed point, and a straight line OPQ revolving round O
cuts a fixed circle in P, Q. On this line is a point R such that OP .OR = k^.
Find the locus of R.
CHAPTER IX.
MISCELLANEOUS PROPERTIES OF THE CIRCLE.
Section I. Orthogonal Circles.
Definition. The angles at which two curves intersect
are the angles between the tangents to the curves at their point
of intersection.
Ex. 318. If two circles intersect at P and Q, the angles at whicli
they intersect at P are equal to the angles at which they intersect
at Q.
Definition. When two circles intersect at right angles, they
are said to intersect orthogonally and are called orthogonal
circles.
Theorem 34.
If two circles are orthogonal, a tangent to either at
their point of intersection passes through the centre of
the other.
The proof is left to the reader.
Ex. 310. Prove Theorem 34.
Ex. 320. Two circles A and B are orthogonal if the tangent to A from
the centre of B is equal to the radius of B.
ORTHOGONAL CIRCLES 77
Ex. 321. Through two given points on a circle draw a circle to cut the
given circle orthogonally.
Is this always possible ?
Ex. 322. Through a given point on a circle draw a circle of given
radius to cut the given circle orthogonally.
Is this always possible?
Ex. 323. The tangents drawn from a point P to two circles are equal;
prove that a circle can be described with P as centre to cut both circles
orthogonally.
Ex. 324. The pole of the common chord of two orthogonal circles with
respect to one of the circles is the centre of the other.
Theorem 35.
The sum of the squares on the radii of two orthogonal
circles is equal to the square on the distance between
their centres.
The proof is left to the reader.
Ex. 325. Prove Theorem 35.
Ex. 326. State and prove the converse of Theorem 35.
Ex. 327. If two circles be described upon the straight lines joining the
two pairs of conjugate points of a harmonic range as diameters, the circles
cut orthogonally.
Theorem 36.
Any diameter of a circle which cuts an orthogonal
circle is divided harmonically by the orthogonal circle.
The proof is left to the reader.
Ex. 328. Prove Theorem 36.
Ex. 329. If P, Q divide a diameter of a given circle harinonicaUy,
any circle through PQ cuts the given circle orthogonaUy.
Ex. 330. A variable circle passes through a fixed point and cuts a given
circle orthogonally ; prove that the variable circle passes through another
fixed point.
78 THE CIRCLE OF APOLLONIUS
Ex. 331. Describe a circle to cut a given circle orthogonally and pass
through two given points.
Is this always possible ?
Ex. 332. If a pair of orthogonal circles intersect at P and Q, and if the
line APB cuts the circles at A and B, then AB subtends a right angle at Q.
Ex. 333. Circles are orthogonal if the angles in the major segments on
opposite sides of the chord of intersection are complementary.
Ex. 334. The locus of the points of intersection of the straight lines
joining two fixed points on a circle to the extremities of a variable diameter
is the circle through the fixed points orthogonal to the given circle.
Section II. The Circle of Apollonius*.
Theorem 37.
If a point P moves so that the ratio of its distances
from two fixed points Q, R is constant, the locus of P is a
circle.
fig. 45.
For any position of P draw PX, PY, the bisectors of the angle
QPR, to cut QR in X, Y respectively.
Since PX bisects _ QPR,
.*. QX : XR - QP : PR
= the given ratio.
.*. X is a fixed point.
Similarly Y is a fixed point.
* See note on p. 20.
THE CIRCLE OF APOLLONIUS T9
Again, since PX, PY are the bisectors of l QPR,
.'. z. XPY is a right angle.
.*. the locus of P is the circle on XY as diameter.
Ex. 335. Construct a triangle having given its base, the ratio of its
other two sides and its area.
Ex. 336. Construct a triangle having given one side, the angle opposite
to that side and the ratio of the other two sides.
Ex. 337. Find a point such that its distances from three given points
are in given ratios.
How many solutions are there ?
Ex. 338. Gi jn the ratio of <^^e two sides of a triangle, the middle
point of tht / 'r-d ride. "^ ich this side is met by the bisector
of the angle opposit 'ion of this bisector, construct the
triangle.
Ex. 339. In fig, ^5 rove thi; xn^„nt at P passes through the
circu ' itre of the triangle PQR.
Ex. 34 O. The internal and external bisectors of the angles of a triangle
>.re rawn, and on the lengths they intercept on the opposite sides circles are
dt, cri ?d having these intercepts as diameters : prove that these circles will
all pass through two points.
80
PTOLEMY S THEOREM
Section III. Ptolemy's"^ Theorem.
Theorem 38.
The sum of the rectangles contained by opposite sides
of a cyclic quadrilateral is equal to the rectangle con-
tained by its diagonals.
fig. 46.
Let PQRS be the quadrilateral.
Make z_ SPT = ^ RPQ, and let PT cut SQ at T.
Now As SPT, RPQ are equiangular
( L SPT = L RPQ, L PST = L PRQ),
.'. PS : PR = ST : RQ,
.". PS . RQ= PR . ST.
Again As TPQ, SPR are equiangular
( L TPQ= z_ SPR, L PQT = L PRS),
.-. PQ : PR = TQ : SR,
.-. PQ.SR^PR.TQ,
.'. PS. RQ+ PQ. SR =: PR . ST + PR. TQ
- PR . SQ.
Ex. 341. What does Ptolemy's theorem become in the special case in
which two vertices of the quadrilateral coincide ?
Ex. 342. What does Ptolemy's theorem become in the special case in
which the circle becomes a straight line ?
Prove the theorem independently.
Ex. 343. ABC is an equilateral triangle inscribed in a circle ; P is any
point on the minor arc BC. Prove that PA= PB + PC.
* Ptolemy was a great Greek astronomer, and one of the earliest writers
on trigonometry (87 — 165 a.d.).
ptolemy's theorem 81
Theorem 39.
The rectangle contained by the diagonals of a quadri-
lateral is less than the sum of the rectangles contained
by its opposite sides unless the quadrilateral is cyclic, in
which case it is equal to that sum.
S R
fig. 47.
Let PQRS be the quadrilateral.
Make l SPT = l RPQ and l PST = L PRQ.
Now As SPT, RPQ are equiangular by construction,
.'. PS : PR = ST : RQ,
.'. PS. RQ==PR. ST.
Also PT : PQ == PS : PR,
.•. PT : PS = PQ : PR
and :! TPQ = z. SPR,
.'. As TPQ, SPR are equiangular,
.•. PQ : PR==TQ : SR,
.*. PQ. SR- PR . TQ,
.-. PR . ST + PR . TQ = PS . RQ + PQ . SR.
But SQ < ST + TQ unless STQ is a straight line.
.*. PR . SQ < PS . RQ + PQ . SR unless STQ is a straight line.
If STQ is a straight line,
z. QSP= z. QRP by construction.
.'. in that case P, Q, R, S are coney clic.
Note that this theorem includes the converse of Ptolemy's theorem.
G. s. M. G. 6
82 Ptolemy's theorem
Ex. 344. If ABC is an equilateral triangle, find the locus of a point
which moves so that the sum of its distances from B and C is equal to its
distance from A.
Several theorems in trigonometry may be proved by means
of Ptolemy's theorem, but of course the proofs do not apply to
angles greater than two right angles.
As an example, we will prove that
sin (a + /5) = sin a cos p + cos a sin y8.
In fig. 48, let PR be a diameter and z. SPR = a, z. RPQ = y8, and
let p be the radius of the circle.
Then PQ == 2p cos ^, RQ = 2p sin ^, SR = 2p sin a, PS = 2p cos a,
PR - 2p. Also by Th. 5 SQ - 2p sin (a + (S).
By Ptolemy's theorem
PR. SQ- PS . RQ + PQ . SR,
.'. 2p . 2p sin (a + ^) = 2p cos a . 2/3 sin ^ + 2p cos /? . 2p sin a,
.'. sin (a+j8) = cos a sin /8 + cos /? sin a.
Ex. 345. Prove the formula for cos(a + /3) by taking PQ a diameter,
zQPR^a, and z PQS = /3.
Ex. 346. Prove the formulae for sin (a - /3) and cos (a - /3).
CONTACT PROBLEMS 83
Section IV. Contact Problems.
Consider the problem of describing a circle to touch three
given circles. As particular cases any of the three circles may
become a line or point.
For the sake of clearness it will be convenient to adopt abbreviations in
Exs. 347—357, e.g. "Describe a circle having given P2L1C0" will be used
as an abbreviation for "Describe a circle to pass through two given points
and touch a given line."
Ex. 347. Show that ten different cases may arise out of this.
Ex. 348. State two cases which are already familiar. How
many solutions are there in each case?
Ex. 349. Describe a circle having given PaLjCo.
How many solutions are there ?
[Produce the line joining the two points to cut the given line ; where will
the point of contact be ?]
Ex. 350. Describe a circle having given PgLoCi.
How many solutions are there ?
[Draw any circle through the two points to cut the given circle; let
their radical axis meet the line joining the two points in T ; draw tangents
from T.]
Ex. 351. Describe a circle having given P1L2C0.
How many solutions are there 1
[Describe any circle touching the two lines and magnify it.]
Ex. 352. If a circle touches a line and a circle, the line joining the
points of contact passes through one end of the diameter at right angles to
the given line.
Note that the ends of the diameter are the centres of similitude of the
line and circle.
Ex. 353. Describe a circle having given PiLiCi.
How many solutions are there 1
[See Ex. 352 ; let A, B be the ends of the diameter, and let AB cut the
line in C ; let M, N be the points of contact, P the given point, and let AP
cut the required circle in P' ; then AB.AC=:AM.AN=AP.AP'.]
6—2
84 MISCELLANEOUS PROPERTIES OF THE CIRCLE
Ex. 354. Describe a circle having given PiLoCg.
How many solutions are there"?
[Take a centre of similitude of the two circles, and see note to Ex. 353.]
Ex. 355. Describe a circle having given PoLgCi.
How many solutions are there ?
[Move the lines parallel to themselves through a distance equal to the
radius of the circle; describe a circle to touch these lines and pass through
the centre of the given circle ; this circle will be concentric with the required
circle.
This process is called thie metliod of parallel translation.]
Ex. 356. Describe a circle having given PoLjCa.
How many solutions are there ?
[Reduce one of the circles to a point by the method of parallel translation.]
Ex. 357. Describe a circle having given P0L0C3.
[Use the method of parallel translation.]
Exercises on Chapter IX.
Ex. 358. Prove that the locus of the centres of circles passing through
a given point and cutting a given circle orthogonally is a straight line.
Ex. 359. Show that, if AB is a diameter of a circle which cuts two given
circles orthogonally, the polars of A with respect to the two circles intersect
in B.
Ex. 360. O is a common point of two orthogonal circles, A, A' are the
points of contact of one common tangent, B, B' of the other.
Show that one of the angles AOA', BOB' is half a right angle and that
their sum is two right angles.
Ex. 361. Two fixed circles intersect in A, B ; P is a variable point on
one of them; PA meets the other circle in X and PB meets it in Y. Prove
that BX and AY intersect on a fixed circle.
Ex. 362. Find the locus of the points at which two given circles sub-
tend the same angle.
MISCELLANEOUS PROPERTIES OF THE CIRCLE 85
Ex. 363. If A, B be two fixed points in a fixed plane, and P a point
which moves in the plane so that AP — w. BP, where m>l, show that P
describes a circle whose radius is ' , .
m-^- 1
Show also that if two tangents to the circle be drawn from A, their
chord of contact passes through B.
Ex. 364. Four points A, B, A', B' are given in a plane; prove that there
are always two positions of a point C in the plane such that the triangles
CAB, CA'B' are similar, the equal angles being denoted by corresponding
letters.
Ex. 365. Three chords AA', BB', CC of a circle are concurrent. Show
that the product of the lengths of the chords AB', BC, CA' is equal to that
of the chords BA', CB', AC.
Ex. 366. Show that a line cannot be divided harmonically by two
circles which cut orthogonally, unless it passes through one or other of
the centres.
Ex. 367. The bisectors of the angles A, B, C of a triangle cut the
opposite sides in Xj, X^; Yj, Y.^ ; Zj, Z.^ respectively.
Show that the circles on the lines X^Xa, Y^Yg, Z^Zg as diameters have
a common chord.
Ex, 368. Construct a triangle, having given the length of the internal
bisector of one angle, the ratio of the side opposite that angle to the sum of
the other sides, and the difference of the other angles.
Ex. 369. It is required to draw a circle to touch two given straight lines
and a given circle. Prove that the eight possible points of contact with the
circle may be found thus : —
Draw tangents to the circle parallel to the two lines and join the vertices
of the rhombus so formed to the point of intersection of the two lines.
These lines cut the circle in the required points.
Ex. 370. Describe a circle :
(i) to touch a given line and pass through two given points,
(ii) to pass through two given points and cut off from a given line
a chord of given length,
(iii) to pass through two given points, so that the tangent drawn to it
from another given point may be of given length.
Ex. 371. Two circles, centres A and B, intersect at right angles at Q
and Q'. A line PQR cuts the circles again at P and R. Show that AB sub-
tends a right angle at the middle point of PR.
86 MISCELLANEOUS PROPERTIES OF THE CIRCLE
Ex. 372. From a given point O, straight lines OA, OB, OC are drawn
cutting a fixed straight line in A, B, C. A circle OBD is described cutting
the circle OAC orthogonally, D being a point on the straight line ABC.
Prove that either the angles AOBjand COD are complementary, or one of
these angles and the supplement of the other are complementary.
Ex. 373. On a given chord AB of a circle, a fixed point C is taken, and
another chord EF is drawn so that the lines AF, BE, and the line joining C
to the middle point of EF meet in a point O ; show that the locus of O is a
circle.
Ex. 374. A straight line is drawn cutting the sides BC, CA, AB of the
triangle ABC in the points D, E, F respectively, so that the ratio FD to DE
is constant; show that the citcles FBD, CDE pass through a fixed point.
Ex. 375. If S, S' are the centres of similitude of two circles, prove that
the circles subtend equal angles at any point on the circle whose diameter is
SS'.
Ex. 376. Construct a quadrilateral given the two diagonals, the angle
at which they cut, and a pair of opposite angles.
Ex. 377. A variable circle passes through a fixed point C and is such
that the polar of a given point A with respect to it passes through a fixed
point B ; show that the locus of the centre of the circle is a straight line
perpendicular to that joining C to the middle point of AB.
Ex. 378. If two sides of a triangle of given shape and size always pass
through two fixed points, the third side always touches a fixed circle.
[The centre of this circle lies on the locus of the vertex of the triangle,
and its radius is equal to an altitude of the triangle.]
Ex. 379. If two sides of a triangle of given shape and size slide along
two fixed circles, the envelope of the third side is a circle. [Bobillier's
Theorem.]
CHAPTEH X.
THE RADICAL AXIS; COAXAL CIRCLES.
Ex. 380. Draw a pair of circles intersecting at points P and Q ; from
any point on PQ produced draw tangents to- the circles ; prove that these
tangents are of equal length.
Definition. The locus of the points from which tangents
drawn to two circles are equal is called the radical axis of the
two circles.
In Ex. 380, we have seen that in the case of two intersecting circles any
point on their common chord produced is on their radical axis. This is a
particular case of the following theorem.
Theorem 40.
The radical axis of two circles is a straight line.
fig. 49.
88
THE RADICAL AXIS
fig. 50.
fig. 51.
[See also Jig. 49 on page 87.]
A, B are the centres of the two circles.
Let P be any point on their radical axis.
Draw PQ, PR tangents to the circles, and draw PN ± to AB.
Since P is on the radical axis
PQ2 = pr2^
.-. AP2 - AQ2 = BP2 - BR2,
.'. PN^ + AN^- AQ^- PN^ + BN^- BR^
.-. AN^ - BN^ = AQ2 - BR2.
.*. having regard to sense
(AN + NB) (AN - NB) = AQ^ - BR^,
.-. AB(AN- NB)=AQ2_ br2,
.'. AN — NB is independent of the position of P,
.'. N is a fixed point.
But PN is ± to AB.
.*. the locus of P is a fixed straight line J_ to AB cutting AB
at a point N, such that AN^ - BN^ = AQ^ _ brI
If we forget this last relation, it is at once recovered from the fact that
the tangents from N are equal.
Note that in the case of intersecting circles the common chord is not,
according to the above definition, part of the radical axis. The following
exercise suggests a modification of the definition which would enable us to
remove this limitation, and regard the whole line as the radical axis.
THE RADICAL AXIS 89
Ex. 381. Prove that if from P, any point on the radical axis of two
circles, lines are drawn cutting the one circle in W, X and the other in
Y, Z, then PW.PX = PY.PZ. Take care that your proof applies to the
common chord of two intersecting circles.
Ex. 382. In the case of each of the following pairs of circles, find the
ratio in which their radical axis cuts the line of centres. Make rough
sketches of the figures. (R, r are the radii of the circles, and d the distance
between their centres.)
(i)
R = 5,
r = 3, d = lQ
(ii)
R = 5,
r = 3, d = 8;
(iii)
R=:5,
T = 3, d = &'.
(iv)
R = 5,
r=3, d=.2'.
(V)
R = 5,
r=3, d = l;
(vi)
R = 5,
r=3, d = Q',
(vii)
R = 5,
r = 0, d = l;
(viii)
R = 5,
r=0, d = 3;
(ix)
R = 5,
r=0, d = Q;
(x)
R=.0,
r = 0, d = b.
Ex. 383. What is the radical axis of two circles, one of which has an
infinite radius (i) when they cut one another, (ii) when they do not cut ?
Ex. 384. What is the radical axis of two circles which touch one
another ?
Ex. 385. The radical axis of two circles bisects their common tangent.
Ex. 386. Suggest a construction for the radical axis of two non-
intersecting circles. (See Ex. 385.)
In what case does the construction fail ?
Ex. 387. In a triangle ABC, the radical axis of its in-circle and the
ex-circle opposite A bisects BC and cuts AB and AC at points whose
distances from A are each equal to \(h + c).
Ex. 388. If the radical axis of the ex-circles opposite A and B cut
AB, AC in X, Y respectively, find the distances AX, AY.
Ex. 389. Three circles pass through the same two points. Prove that,
if the common tangent of any two of them is cut by the third circle, it
is divided harmonically.
Ex. 390. Prove the validity of the following construction for the
radical axis of two circles. Draw any circle to cut the one circle
in Q, Q! and the other in R, R'; produce QQ' and RR' to cut at P;
draw PN i. to the line of centres. Then PN is the radical axis.
90 THE RADICAL AXIS
Ex. 391. What is the radical axis of :
(i) a point-circle and a circle,
(ii) two point-circles,
(iii) a circle and a line (a circle of infinite radius),
(iv) a point- circle and a line,
(v) two concentric circles,
(vi) two parallel lines,
(vii) two intersecting lines ?
Ex. 392. Give a construction for the radical axis of a circle and a point
analogous to the construction of Ex. 390.
Does your construction hold if the point is inside the circle ?
Ex. 393. If from any point P tangents are drawn to two circles, the
difference of their squares is equal to twice the rectangle contained by
the distance between the centres and the perpendicular from P on the
radical axis of the circles.
[Join P to the centres of the circles, and from P draw a perpendicular to
the line of centres.]
Theorem 41.
The three radical axes of three circles taken in pairs
are concurrent.
The proof is left to the reader.
Ex. 394. Prove Theorem 41.
[Consider the tangents from the point where two of the radical axes
intersect.]
Definition. The point of concurrence of the three radical
axes of a system of three circles is called the radical centre of
the three circles.
Ex. 395. If each of three circles touches the other two, the three
common tangents at their points of contact are concurrent.
Ex. 396. Circles are described with the sides of a triangle ABC as
diameters, where is their radical centre?
[What are their radical axes ?]
Ex. 397. Where is the radical centre of three point-circles ?
COAXAL CIRCLES 91
Ex. 398. If the centres of three circles are coUinear, where is their
radical centre?
Ex. 399. Where is the radical centre of three circles, two of which are
concentric ?
Coaxal Circles.
Ex. 400. Draw a circle A and a circle B to touch it ; what is their
radical axis? Find another circle C such that A and C have the same
radical axis as A and B.
Ex. 401. Draw two intersecting circles A and B. What is their radical
axis ? Draw another circle C such that A and C have the same radical axis
as A and B. What is the radical axis of B and C ?
Ex. 402. Draw a circle with centre A and a line PN outside it; draw
AN 1 to PN ; from P draw PT a tangent to the circle; from P draw a line
PT'=:PT, draw a circle with its centre on AN (or AN produced) to touch
PT' at T'. What is the radical axis of the two circles ?
Definition. If a system of circles is such that every pair has
the same radical axis, the circles are said to be coaxal.
It is obvious that coaxal circles have all their centres on a
straight line, which is perpendicular to the common radical axis.
Q
FTTb
fig. 52.
In Theorem 40 it was proved that if A, B are the centres
of two circles whose radii are AQ, BR and N the point at which
their radical axis cuts AB, then
AN2-AQ2 = BN^- BR2.
By reversing the steps of that theorem we could prove that,
if the given relation is true and if tangents are drawn to the
circles from any point P on the perpendicular to AB through N,
these tangents must be equal ; in fact, that if the relation holds
PN is the radical axis of the two circles.
92
COAXAL CIRCLES
Now suppose that the one circle (centre A) and the radical
axis are given ; by taking different positions for B on the line AN
(produced both ways) and choosing in each instance the radius
given by the above relation, we get an infinite number of circles?
the tangents to which from any point P on PN are equal to one
another.
We therefore get a system of coaxal circles.
Intersecting coaxal circles. If any circle of a coaxal
system cuts the radical axis at C and D say, all the circles must
pass through C and D, for the tangent to the one circle from C
(or D) is of zero length, and therefore the tangent from C (or D)
to each circle of the system must be of zero length.
In the same way, if any two circles of the system intersect
at C and D, all the circles must pass through C and D, and CD is
their common radical axis.
This suggests an easy construction for a system of intersecting
coaxal circles.
fig. 53.
COAXAL CIRCLES 93
Non-intersecting coaxal circles. We will now consider
a construction for a system of coaxal circles for the case in
which no circle of the system cuts the radical axis (and no two
circles of the system cut one another).
[See Jig. 55 on page 94.]
fig. 54.
Suppose we have the radical axis and one circle of the
system.
From N (which must be outside all the circles) draw a
tangent NQ to the circle.
With centre N and radius NQ describe a circle.
Draw BR a tangent to this circle from any suitable point
in AN (or that line produced). Then the circle with B as centre
and BR as radius will be a circle of the system.
For AN^-AQ^:^ NQ2:=:NR2=BN2-BR2.
It should be noticed that instead of taking N as centre we might take
any point on the radical axis. This method would then apply to inter-
secting circles as well as non-intersecting.
Ex. 403. Draw a system of coaxal circles, one circle of the system
having its centre 4 cm. from the radical axis and having a radius of 3 cm.
It is worthy of special notice that in a system of coaxal
circles one member of the system consists of the radical axis and
the line at infinity.
Ex. 404. In fig. 54 what position of R will give the radical axis as a
member of the system ?
94
COAXAL CIRCLES
fig. 55.
Ex. 405. From what points of the line AB in fig. 54 is it impossible
to draw tangents to the construction circle ?
Take a point B' between L and N ; according to the formula, what would
be the square on the radius of the circle of the system with centre B' ? Is
this positive or negative ?
Ex. 406. What is the radius of the circle of the system whose centre
is at L, where the construction circle cuts AN ?
Limiting Points. It is obvious from the method of con-
structing non-intersecting coaxal circles (and also from the
relation AN^- AQ^^^ BN^- BR^) that B cannot be within the
construction circle, but may be anywhere else along the line
through A and N.
The circles of the system whose centres are at the points L
and L' where the construction circle cuts the line AN have zero
COAXAL CIRCLES 95
radius, i.e. are point circles. L and L' are called the limiting
points of the system.
Definition. The limiting points of a system of coaxal circles
are the point circles of the system.
A system of intersecting coaxal circles has no real limiting
points ; for any point in the line of centres may be taken as the
centre of a circle of the system.
Or, looking at the question from the point of view of the
definition, in the case of intersecting circles there are no real
point circles of the system, for BN^ — BR^^AN^-AQ^ which is
BR^ cannot be zero.
Ex. 407. P is any point on the radical axis of a coaxal system of
circles ; with P as centre a circle is described to cut one of the circles ortho-
gonally ; what is its radius ? Prove that this circle cuts all the circles of the
system orthogonally.
Ex. 408. In Ex. 407 suppose the system to be of the non -intersecting
type ; prove that the orthogonal circle passes through two points which
are the same whatever position on the radical axis is chosen for P.
Ex. 409. In Ex. 407 suppose the system to be of the non-intersecting
type ; prove that an infinite number of circles can be drawn to cut all the
circles orthogonally, and prove that these cutting circles are themselves
coaxal.
Theorem 42.
With every system of coaxal circles there is asso-
ciated another system of coaxal circles, and each circle
of either system cuts every circle of the other system
orthogonally.
Since the tangents to a system of coaxal circles (A) from any
point P on their radical axis are equal to one another, it follows
that the circle (B) with centre P and any one of these tangents
as radius will cut all the circles of the system (A) orthogonally.
yb COAXAL CIRCLES
Again, since there is an infinite number of positions of P on
the radical axis, there is an infinite number of circles (B) each of
which cuts all the circles of the system (A) orthogonally.
We have still to show that these cutting circles (b) form
another coaxal system.
Consider any one circle of the system (A) ; the tangents from
its centre to the orthogonal circles (B) are each a radius of the
(a) circle, and therefore equal to one another; similarly for any
other circle of the system (A).
.'. the orthogonal circles (B) are coaxal, their radical axis
being the line of centres of the system (A).
fig. 56.
COAXAL CIRCLES 97
Theorem 43.
Of two orthogonal systems of coaxal circles, one
system is of the intersecting type and the other of the
non-intersecting type, and the limiting points of the
latter are the common points of the former.
Suppose that a system (A) of coaxal circles is of the non-
intersecting type and has limiting points L and L' ; since L and L'
are point circles of the system, it follows that all the circles of
the orthogonal system (B) pass through L and L'> and therefore
that the system (B) is of the intersecting type, L and L' being the
common points.
Again suppose the given system is of the intersecting type,
M and M' being the common points ; we see that no circle of the
orthogonal system can have its centre between M and M'; there-
fore these are the limiting points of the orthogonal system.
Ex. 410. Draw a system of coaxal circles which touch one another;
draw the orthogonal system. Where are their Hmiting points and common
points ?
Ex. 411. The radical axes of a given circle and the circles of a coaxal
system are concurrent.
Ex. 412. Prove that a common tangent to any two circles is divided
harmonically by any coaxal circle which cuts it.
Ex. 413. If L is one of the limiting points of a system of coaxal circles
and XLY is any chord of a circle of the system, the distances of X, L, Y
from the radical axis are in geometrical progression.
Ex. 414. A common tangent to any two circles of a non-intersecting
coaxal system subtends a right angle at each of the limiting points.
Ex. 415. Tlie polar of eitlier limiting point of a coaxal system
with regard to any circle of the system passes through the other
limiting point.
Ex. 416. The tangents to a family of circles from a point A are all
equal to one another; and the tangents from another point B are also equal
to one another; prove that the circles are all coaxal. What is the condition
that the system should be of the non-intersecting type, and what are the
limiting points in that case ?
G. S. M. G. . 7
98 RADICAL axis; coaxal circles
Ex. 417. Prove that the polars of a fixed point P with regard to a
system of coaxal circles with real limiting points all pass through another
fixed point, namely that point on the circle through P and the limiting
points which is at the other extremity of the diameter through P.
Exercises on Chapter X.
Ex. 418. If T be a point from which equal tangents can be drawn to
two circles whose centres are A and B, prove that the chords of contact
of tangents from T intersect on the line through T at right angles to AB.
Ex. 419. The mid-points of the four common tangents to two non-
intersecting circles are collinear.
Ex. 420. If each of three circles intersects the other two, prove that
their common chords are concurrent.
Ex. 421. Three circles, centres D, E, F, touch each other two and two
in A, B, C. Prove that the circumcircle of ABC is the in-circle of DEF.
Ex. 422. Show how to describe a circle of a given coaxal system to
touch another given circle (i) when the system is of the intersecting, (ii) of
the non-intersecting type.
Ex. 423. Consider the various Apollonius' circles for two fixed points
obtained by varying the given ratio ; are they coaxal ?
Ex. 424. If a system of circles have the same polar with regard to a
given point, show that they are coaxal, and find the position of the common
radical axis.
Ex. 425. Prove that the four circles whose diameters are the common
tangents to two non-intersecting circles have a common radical axis.
Ex. 426. Show that the limiting points of a pair of non-intersecting
circles and the points of contact of any one of their common tangents lie on
a circle cutting the two circles orthogonally.
Ex. 427. The circle whose diameter is the line joining the centres of
similitude of two circles is coaxal with those circles.
Ex. 428. If two circles X and Y cut orthogonally, prove that the polar
with respect to X of any point A on Y passes through B, the point diametri-
cally opposite to A.
If the polars of a point, with respect to three circles whose centres are
on a straight line, are concurrent, prove that the three circles are coaxal.
COAXAL CIRCLES 99
Ex 429. Prove that the common orthogonal circle of three given circles
is the locus of a point whose polars with respect to the three circles are
concurrent.
Ex. 430. The external common tangent to two circles which lie outside
one another touches them in A and B ; show that the circle described on
AB as diameter passes through the limiting points L and L' of the coaxal
system to which the circles belong.
If O is the mid-point of the above common tangent, prove that OL, OL'
are parallel to the internal common tangents of the circles.
Ex. 431. The internal and external bisectors of the angles of a triangle
are drawn, and on the lengths they intercept on the opposite sides circles
are described having these intercepts as diameters ; prove that these circles
all pass through two points which are collinear with the circumcentre of
the triangle.
Ex. 432. Describe a circle which shall pass through two given points
and bisect the circumference of a given circle.
Ex. 433. Prove that all the circles which bisect the circumferences of
two given circles pass through two common points.
Ex. 434. ABC is a triangle and two circles are drawn, one to touch
AB at A and to pass through C, the other to touch AC at A and to pass
through B. If the common chord of these circles meets BC in D, prove
that BD : DC = BA2: AC2.
Ex. 435. A line PQ is drawn touching at P a circle of a coaxal system
of which the limiting points are K, K', and Q is a point on the line on the
opposite side of the radical axis to P ; show that, if T, T' be the lengths
of the tangents drawn from P to the two concentric circles of which the
common centre is Q and radii are respectively QK, QK', then
T : T' : : PK : PK'.
Ex. 436. The tangents at A, A' to one given circle cut a given non-
intersecting circle in P, Q and P', Q' respectively, and AA' cuts PP' in X.
Show that, if O is a limiting point of the coaxal system determined by the
two given circles, then will OX be a bisector of the angle POP'.
7—2
CHAPTER XI.
INVERSION.
Definition. If O is a fixed point and P any other point, and
if a point P' is taken in OP (produced if necessary) such that
OP. OP' = A;^ (where ^ is a constant), P' is said to be the inverse
of P with regard to the circle whose centre is O and radius k.
O is called the centre of inversion, the circle is called the
circle of inversion, and k the radius of inversion*.
From the definition it is obvious that P is the inverse of P'. Also that
OP' varies inversely as OP ; hence the name.
If P is made to describe any given figure and if P' always
moves so that it is the inverse of P, P' describes a figure which is
called the inverse of the given figure with respect to the circle of
inversion.
When a large number of inverse points have to be found the following
construction is useful.
Assume the above notation ; draw a circle of any radius and a tangent to
it at any point A; from the tangent cut oif a length Ao = the radius of
inversion ; find a point p on the circle such that op — OP ; let p' be the other
point at which op cuts the circle, then op' = OP'.
Ex. 437. What is the inverse of a straight line when the centre of in-
version is on the straight line ?
Ex. 438. What is the inverse of a given circle when the centre of in-
version is the centre of the given circle ?
* Sometimes k'^ is called the constant of inversion.
INVERSION , , ,o>J,01.
Ex. 439. Draw a straight line and mark a point O 4 inches from the
line ; taking O as centre and a radius of inversion 3 inches, mark a number
of points on the inverse of the straight line.
Ex. 440. Draw a circle of radius 2 inches ; take a point O 1 inch from
its centre ; taking O as centre and 1 inch as radius of inversion, plot the
inverse of the circle.
Ex. 441. Draw a circle of radius 2 inches; take a point O 3 inches
from its centre ; taking O as centre and 2 inches as radius of inversion, plot
the inverse of the circle.
Ex. 442. Plot a parabola and invert it (i) with the focus as centre of
inversion, (ii) with the vertex as centre of inversion.
Theorem 44.
If a figure is inverted first with one radius of inver-
sion and then with a different radius, the centre being
the same in both cases, the two inverse figures are
similar and similarly situated, the centre being their
centre of similitude.
If Pi is the inverse of a point P when ky^ is the radius of in-
version and Pg the inverse of the same point when A:^ is the
radius, the centre O being the same in both cases, then
Hence the theorem is true.
In consequence of this property it is generally unnecessary to
specify the radius of inversion; in fact, it is usual to make no
reference to the circle of inversion and to speak of inverting with
regard to a point.
Sometimes we take a negative constant of inversion ; in this
case the circle of inversion must of necessity be avoided as it has
an imaginary radius.
102
INVERSION
Theorem 45.
The inverse of a straight line, with regard to a point
on it, is the line itself.
This is obvious from the definition.
Ex. 443. What is the inverse of a point on the Hne which is infinitely
close to the centre of inversion ?
Ex. 444. What is the inverse of the line at infinity ?
Ex. 445. OABC is a straight line, and A', B', C are the inverses of
A, B, C, when O is the centre of inversion ; if B is the mid-point of AC,
prove that O, A', B', C is a harmonic range.
Ex. 446. If a harmonic range is inverted with regard to any point on
the line, another harmonic range is obtained.
Ex. 447. Prove that Ex. 445 is a particular case of Ex. 446.
Theorem 46.
The inverse of a straight line, with regard to a point
outside it, is a circle through the centre of inversion.
fig. 57.
Let PA be the given line and O be the centre of inversion.
Draw OA i. to PA.
INVERSION 103
Take A', P' the inverses of A, P.
Then OP. OP'=:OA. OA',
.*. P, A, A', P' are concyclic,
.-. Z.OP'A' = ^OAP
= a rt. L.
But O and A' are fixed points.
.*. as P moves along the line PA, P' describes a circle on OA as
diameter.
Ex. 448. Show that Theorem 45 is not really an exception to the
theorem that the inverse of a straight line is a circle through the centre of
inversion.
Ex. 449. Draw a figure showing the inverse of a straight line with
regard to a point outside it for a negative constant of inversion.
Theorem 47.
The inverse of a circle with regard to a point on its
circumference is a straight line at right angles to the
diameter through the centre of inversion.
The proof is left to the reader.
Ex. 4 so. Prove Theorem 47.
Ex. 451. If a circle is inverted with regard to a point on it, the
centre of the circle inverts into the image of the centre of inversion in
the resulting straight line.
Ex. 462. A straight line meets a circle a in the points A, B and a
circle /3 in the points O, D ; O is a point on the radical axis of a and j3.
OA, OB meet a again at A', B' and 00, OD meet j8 again at C, D'.
Show that O, A', B', C, D' lie on a circle.
104 INVERSION
Ex. 453. In fig. 58, OA = OB, AP=rPB = BP'=P'A;
(i) prove that OPP' is a straight line ;
(ii) prove that, if O be fixed, P and P' will move so that they are
inverse points with regard to O.
fig. 58.
Ex. 454. If, in Ex. 453, C is a fixed point, and P moves so that
CP = CO, prove that the locus of P' is a straight line.
Feaucellier's Cell*. Fig. 58 suggests a mechanical device, called a
linkage, for constructing the inverse of a given figure ; a model can be con-
structed consisting of rods freely hinged at the points O, A, B, P, P' ; from
Ex. 453 we see that if O is fixed and P moved along a given curve P'
describes the inverse curve.
Ex. 454 shows that, if P is made to describe a circle through O, P'
moves on a straight line. Now it is not necessary that the rods should be
straight ; the only essential is that the distance between the points O and A
should equal that between O and B, etc., and the equality of these distances
can be tested by superposition. Thus this linkage enables us to draw a
straight line without presupposing that we have a straight edge.
* This linkage was invented in 1873 by Peaucellier, a captain in the
French army.
inversion 105
Theorem 48.
The inverse of a circle with regard to a point not on
its circumference is another circle.
fig. 59. fig. 60.
Let O be the centre of inversion.
Draw a line OPQ to cut the circle at P and Q.
Let P' be the inverse of P.
Then OP . OP' = constant.
But OP . OQ =: constant,
.'. op' : OQ = constant.
But the locus of Q is a circle,
.*. the locus of P' is a circle.
I.e. the inverse of the given circle is a circle.
Note that in figs. 59 and 60 the parts of the circles which are thickened
are inverses of one another.
Ex. 455. Show how to invert a circle into itself, the centre of inversion
being outside the circle.
Ex. 456. Is it possible to invert a circle into itself (i) with regard to
a point inside the circle, (ii) with regard to a point on the circle ?
Ex. 457. Show how to invert simultaneously each of three circles into
itself.
Ex. 458. If a circle is inverted witli regard to any point not on
its circumference, its centre inverts into the point at which the line
of centres cuts the polar of the centre of inversion with respect to
the inverse circle.
Ex. 459. Show that Ex. 451 is a particular case of Ex. 458.
106
INVERSION
Theorem 49.
Two curves intersect at the same angles as their
inverses.
fig. 61.
Let P be a point of intersection.
Through O, the centre of inversion, draw a straight line
making a small angle with OP to cut the curves in Q and R.
Let P', Q', r' be the inverses of P, Q, R respectively.
Join PQ, PR, PR', p'Q'.
Since OP. OP' = OQ. OQ',
.*. P, P', Q', Q are concyclic,
.*. ^OPQ=Z.OQ'P'.
Similarly lOPR = l OR'P',
.*. L QPR = L R'P'Q'.
Now, as OQ moves up to OP, so PQ, PR, P'Q', P'r' move up to
and in the limit coincide with the tangents to the curves at P
and P'.
.'. the angles between the tangents at P are equal to the
angles between the tangents at P'.
.*. two curves cut at the same angles as their inverses.
Ex. 460. Give an independent proof of Theorem 49 in the case of two
straight lines inverted into a straight line and a circle.
Ex. 461. Give an independent proof of Theorem 49 in the case of two
straight lines inverted into two circles.
Ex. 462. Prove that the tangent to a curve from the centre of inver-
sion is also a tangent to the inverse curve.
INVERSION
107
By applying the above results we can deduce new theorems
from theorems we know already; this process is called inverting
a theorem.
Example I.
Invert the following theorem with regard to the point O : —
If O, A, B, C are four points on a circle, angles AOC, ABC are
equal or supplementary.
Let a', B", C' be the inverses of A, B, C.
We will write the corresponding properties of the figure and
its inverse in parallel columns.
[It is convenient to draw the figures separately.]
fig. 62.
fig. 63.
OABC is a circle,
OA, OC are st. lines,
AB is a st. line,
BC is a St. line,
L AOC is equal or supplemen-
tary to L ABC.
A'b'C' is a straight line,
OA', OC' are st. lines,
OA'B' is a circle,
OB'C' is a circle,
L A'OC' is equal or supplemen-
tary to L at which circles
OA'B', OB'C' intersect.
Hence we deduce the theorem that, if A'B'C' is a straight line,
and O a point outside it, the angles at which the circles OB'A',
OB'C' intersect are equal or supplementary to the angle A'OC'.
108
INVERSION
Example II.
Prove the following theorem by inverting with regard to the
point O. AOBF, AOCE are two circles intersecting at O, A; FO a
diameter of the first cuts the second at C, EO a diameter of the
second cuts the first at B; then AO passes through the centre
of the circle OBC.
Let A', B'
be the inverses of A. B,
We will now write the corresponding properties of the figure
and its inverse in parallel columns.
fig. 65.
fig. 64.
AOBF, AOCE are two circles
through A, O,
FO, a diameter of 0AOBF, cuts
0AOCE at C;
EO, a diameter of 0AOCE, cuts
©AOBF at B.
To prove that AO passes through
the centre of 0OBC.
A'B'F', A'C'E' are two st. lines
through A',
F'O, the perpendicular from O
on A'b'F', cuts A'C'E' at C';
E'O, the perpendicular from O
on A'C'E', cuts A'b'F' at B'.
To prove that AO is perpen-
dicular to B'C'.
Now we see that the inverse theorem is true (it is the ortho-
centre property of a triangle).
the original theorem is true.
INVERSION 109
Ex. 463. Invert the following theorem with regard to the point O : If
O, A, B, C are four points on a circle, angles OAC, OBC are equal or
supplementary.
Ex. 464. Invert the theorem 'The angle in a semicircle is a right angle *
with regard to one end of the diameter.
Ex. 465. OP and OQ are lines through a fixed point O, inclined at a
constant angle and intersecting a fixed line in P, Q ; the envelope of the
circle round OPQ will be another circle.
Ex. 466. Prove by inversion (or otherwise) that if the circumcircles of
two triangles ABC, ABD cut orthogonally, then the circumcircles of CAD
and CBD also cut orthogonally.
Ex. 467. Prove by inversion that the circles having for diameters three
chords OA, OB, OC of a circle intersect again by pairs in three coUinear
points.
Ex. 468. Three circles OBC, OBE, OCF pass through a point O; OBF
is a straight line passing through the centre of the circle OCF; OCE is a
straight line passing through the centre of the circle OBE; prove that
circles OBE, OCF intersect on OD the diameter through O of the circle
OBC.
Ex. 469. Prove by inversion that a straight line drawn through a point
O to cut a circle is divided harmonically by the circle and the polar of O.
[Invert with regard to O.]
Ex. 470. Tlie limiting points of a coaxal system are inverse
points witli regard to any circle of the system.
Ex. 471. A system of intersecting coaxal circles inverted with
regard to a point of intersection becomes a system of straight lines
through a point.
Ex. 472. Invert the following theorem with regard to the point O: If
each of a system of circles passes through two given points O and O',
another system of circles can be described which cut the circles of the first
system orthogonally.
Ex. 473. A system of non -intersecting coaxal circles inverted
with respect to a limiting point of the system becomes a system of
concentric circles having the inverse of the other limiting point for
centre.
[Consider the orthogonal system of circles and use Ex. 472.]
110 INVERSION
Ex. 474. What is the inverse of a system of intersecting coaxal circles
with respect to any point ?
Ex. 475. What is the inverse of a system of non-intersecting coaxal
circles with respect to any point ?
Inversion may be applied to geometry of three dimensions.
By rotating the figures of theorems 46, 47, 48 about the line
through the centre of inversion and the centres of the circles we
arrive at the following results :
(i) The inverse of a plane with regard to a point outside it
is a sphere through the centre of inversion.
(ii) The inverse of a sphere with regard to a point on its
surface is a plane at right angles to the diameter through the
centre of inversion.
(iii) The inverse of a sphere with regard to a point not on its
surface is another sphere.
Ex. 476. What is the inverse of a circle with regard to a point not in
its plane ?
[Regard the circle as the intersection of a sphere and a plane.]
Ex. 477. A circle is inverted with respect to a sphere whose centre O
does not lie in the plane of the circle ; prove that the inverse is a circle, and
show that the point P which inverts into the centre of the inverse circle is
obtained thus : Describe a sphere through O and the circumference of the
given circle ; then P is the pole of the plane of the circle with respect to
this sphere.
Exercises on Chapter XL
Ex. 478. Q is the inverse of P with respect to a circle whose centre
is O, AQB is any chord of the circle ; prove that PQ bisects the angle APB.
Ex. 479. A circle, its inverse, and the circle of inversion are coaxal
with one another.
Ex. 480. Show that it is possible to invert three circles so that the
centres of the inverse circles are collinear.
INVERSION 111
Ex. 481. If two circles cut orthogonally the inverse of the centre of the
first with respect to the second coincides with the inverse of the centre of
tbe second with respect to the first.
Ex. 482. Two points are inverse with respect to a circle ; show that, if
the figure be inverted with respect to any circle, the new figure will have the
same property.
Ex. 483. A, B,.C are three points in a straight line and P any other
point. AFE, BFD, CED are drawn perpendicular to PA, PB, PC respectively ;
prove that P, D, E, F lie in a circle.
Obtain a new theorem by inverting with respect to P.
Ex. 484. If c is the distance between the centres of two intersecting
circles whose radii are r, r', show that the ratio c^-r^-r'^ : rr' is unaltered
by inversion with regard to any point external to the two circles.
Ex. 485. Two circles intersect at O and P and their tangents at O meet
the circles again at A and B. Show that the circle circumscribing the
triangle AOB cuts OP produced at a point Q such that 0Q=20P, and that
if a line is drawn through P parallel to the tangent at O to the circle AOB,
then the part of this line intercepted between OA and OB is bisected at P.
Ex. 486. If A, B^ C be three collinear points and O any other point,
show that the centres P, Q, R of the three circles circumscribing the triangles
OBC, OCA, CAB are concyclic with O. Also that if three other circles are
drawn through O, A ; O, B ; O, C to cut the circles OBC, OCA, CAB,
respectively, at right angles, then these three circles will meet in a point
which lies on the circumcircle of the quadrilateral OPQR.
Ex. 487. From any point P on the circle ABC a pair of tangents PQ,
PR are drawn to the circle DEF and the chord QR is bisected in S. Show
that the locus of S is a circle ; except when the circle ABC passes through
the centre of the circle DEF, in which case the locus of S is a straight line.
Ex. 488. Through one of the points of intersection of two given circles
any line is drawn which cuts the circles again in P, Q respectively. Prove
that the middle point of PO is on a circle whose centre is midway between
the centres of the given circles.
Ex. 489. Show that there is in general one circle of a coaxal system
which cuts a given circle orthogonally.
What is the exceptional case?
Ex. 490. Show that circles which cut one given circle orthogonally and
another given circle at a given angle will also cut a third fixed circle at the
same fixed angle.
112 INVEKSION
Ex. 491. A, B, C, D are four coplanar points. Prove that in an infinite
number of ways two circles can be drawn making an assigned angle with
each other, and such that A and B are a pair of inverse points of one circle,
and C and D of the other circle.
Ex. 492. If P', Q' are the inverses of P, Q with respect to a point O,
PQ : P'Q'= OP . OQ : fc2^ where A; 2 is the constant of inversion.
Ex. 493. Invert with respect to the point O the proposition : If PAQ,
RAS are two chords of a circle which passes through O, the rectangle
PA . AR = rectangle RA . AS.
Ex. 494. The sides of a triangle ABC touch a circle whose centre is O,
and on OB, 00 produced, if necessary, are taken points B' and C respec-
tively such that OB . OB' = OC . 00'= 0A2. Prove that O is the orthocentre
of the triangle AB'C.
Ex. 495. Two given circles intersect in a point O ; prove, by the method
of inversion, that the inverse point of O with respect to any circle which
touches them lies on one or other of two fixed circles which cut one another
orthogonally.
Ex. 496. If two circles be inverted with respect to a circle whose centre
is at their external centre of similitude and whose (radius)^ is equal to the
rectangle contained by the tangents to the circles from its centre, prove
that the radical axis of the two circles inverts into the circle on the line
joining the two centres of similitude as diameter.
Ex. 497. Prove that any two circles are inverse to one another with
respect to some third circle ; and that with any point on this third circle as
origin of inversion the two circles will invert into equal circles.
Ex. 498. (i) A sphere is inverted from a point on its surface; show
that to a system of parallels and meridians on the surface will correspond
two systems of coaxal circles in the inverse figure.
(ii) Prove that, if P, Q be the ends of a diameter of a small circle of a
sphere, O a point of the great circle PQ, and R any point on the circle, then
the arcs of the small circles PRO, RQO are perpendicular to each other at R.
Ex. 499. (i) A circle is inverted from a point which is not upon its
circumference and not necessarily in the plane of the circle. Show that the
inverse curve is also a circle.
(ii) Circles are drawn to cut a given circle orthogonally at two points
of intersection and to pass through a given point not in the plane of the
circle. Show that they intersect in another common point ; and hence show
how a circle and a point not in its plane may be inverted respectively into
circle and centre.
INVERSION 113
Ex. 500. Show that the locus of points with respect to which an anchor
ring can be inverted into another anchor ring consists of a straight line and
a circle.
Ex. 501. The figures inverse to a given figure with regard to two circles
Ci and C2 are denoted by Si and S2 respectively ; show that, if Ci and C2
cut orthogonally, the inverse of Si with regard to C2 is also the inverse of 82
with regard to Ci.
Ex. 502. r is a circle and P and Q are any two points inverse to it;
r' , P' , Q' are the respective inverses with regard to any point. Show that
P', Q' are inverse points with regard to the circle V.
Ex. 503. (i) Show that, if the circles inverse to two given circles ACD,
BCD with respect to a point P be equal, the circle PCD bisects (internally
or externally) the angles of intersection of the two given circles.
(ii) Prove that four points P, Q, R, 8 can be found such that with
respect to any one of them the points inverse to four given points A, B, C, D
form a triangle and its orthocentre ; and that the points inverse to P, Q, R, 8
with respect to any one of the four A, B, C, D also form a triangle and its
orthocentre.
Ex. 504. A circle moving in a plane always touches a fixed circle, and
the tangent to the moving circle from a fixed point is always of constant
length. Prove that the moving circle always touches another fixed circle.
G. s. M. Q.
CHAPTER XII.
ORTHOGONAL PROJECTION.
Suppose that we have a plane (say a sheet of glass) with a
variety of figures drawn upon it.
And let this plane be placed, in an inclined position^ above a
second — horizontal — plane.
If a distant light (e.g. the sun) be allowed to shine upon the
figures drawn on the glass, and to cast shadows of them upon
the horizontal plane, these shadows would be ' projections ' of the
original figures.
If the sun is directly overhead, so that its rays strike per-
pendicularly upon the horizontal plane, the projection is called
'orthogonal.'
The definition of orthogonal projection is as follows .
Definition. Let there be an assembly of points (a) in a
plane {p). From each point let a perpendicular be drawn to
a second plane {q). The feet of these perpendiculars together
constitute the orthogonal projection of the assembly {a).
We must now enquire what relations exist between figures
and their orthogonal projections upon other 'planes of projection.'
In what follows, it must be assumed that the projection is
orthogonal unless the contrary is stated or distinctly implied.
1. The projection of a straight line is a straight line.
The perpendiculars from all points on the original line form
a plane, which cuts the plane of projection in a straight line.
ORTHOGONAL PROJECTION
115
2. A point of intersection of two curves in the
original plane projects into a point of intersection of
the resulting curves.
3. A tangent to a curve, and its point of contact,
project into a tangent to the resulting curve and its point
of contact.
4. The lengths of lines are usually altered by orthogonal
projection ; in fact, the lines are foreshortened.
Ex. 505. Take the case of projection on to a horizontal plane from a
plane inclined to it at 60°.
Prove that all the lines of steepest slope are halved by projection.
Are any lines unaltered by projection ?
What is the condition that two lines that are equal before projection shal
remain equal after projection?
If a be the length of a segment of one of the lines of
steepest slope in a plane, and 0 the angle which the plane
makes with the plane of projection, then «cos^ is the
length of the projection of a.
AB is the segment a, CD is its projection.
In the plane AEC draw BF i| to DC, meeting AC in F.
Then ^ ABF = ^ AEC = (9,
.*. DC = BF == acos^.
8—2
116 ORTHOGONAL PROJECTION
5. Lines parallel to the plane of projection are un-
altered in length by projection.
6. If A be an area in a plane, its projection has area
A cos 6.
fig. 67.
Let the area be divided up into strips A BCD by lines of
steepest slope.
By drawing parallels to the plane of projection, cut off a
rectangle AECF from each strip.
Let A'E'C'F' be the projection of AECF.
Now A' F' = AF cos 6, A' E' = AE.
.'. rect. A'E'C'F' = rect. AECF x cos ^.
If the strips become very narrow (and therefore numerous),
then each strip tends to equality with the corresponding rectangle,
the neglected portions being comparatively unimportant ; and it
is shown in the infinitesimal calculus that, in the limit, no error
is made by regarding the area as composed of infinitely narrow
rectangles.
But each rectangle is diminished by projection, in the ratio
cos^ : 1.
.*. the projection has area A cos B.
Ex. 506. Give an independent proof of the above theorem for a triangle,
by drawing through its vertices perpendiculars to the line of intersection of
the planes, and considering the three trapezia thus formed.
Hence prove the theorem for any rectilinear figure.
ORTHOGONAL PROJECTION 117
7. Parallel lines project into parallel lines.
The intersection of the two parallel lines is a point at infinity.
This projects into a point at infinity.
Therefore the two projected lines are parallel.
8. Parallel lines are diminished, by projection, in the
same ratio.
fig. 68.
AB, A'b' are parallel; CD, C'D' are their projections.
Draw AE || to CD, A'E' to C'D'.
Let AE meet BD in E, A'E' meet B'd' in E'.
Now CD is II to C'D' by (7). Thus we have AE || to CD, CD
to C' D', C' D' II to A' E'.
.*. AE is II to A'E'.
Also AB is II to A'B'.
.*. by a theorem in solid geometry
Z.BAE = z.B'A'E', =:<^ (say),
.*. AE = AB cos <;^, A' E' = A' B' cos </>.
But AE = CD, the proj" of AB, and A'E' = C'D', the proj" of
AB'
Therefore the two parallel lines are both diminished in the
same ratio by projection.
118 ORTHOGONAL PROJECTION
9. If a line and any number of points on it be pro-
jected, the projection is divided in the same ratio as the
original line.
This follows from (8). The following particular case is useful.
10. The projection of the mid-point of a line bisects
the projection of the line.
11. It has been seen that a number of geometrical relations
are unaltered by orthogonal projection ; and the beginner may
be tempted to apply this principle too freely.
It must be noted that, as a rule, angle properties are destroyed
by orthogonal projection.
Ex. 607. Discover cases in which a right angle is unaltered by pro-
jection.
Ex. 508. One arm of an angle is |1 to the plane of projection. Is the
angle increased or diminished by projection ?
Ex. 509. One arm of an angle is a line of greatest slope. Is the angle
increased or diminished by projection ?
Ex. 510. Answer the question of Ex. 609 for an angle whose bisector is
(i) a line of greatest slope,
(ii) a parallel to the plane of projection.
Ex. 511. Discover any case in which the relation of an angle and its
bisector is unchanged by projection.
Ex. 512. Prove that the relation of Ex. 511 is not preserved generally,
by considering the particular case of
(i) a right angle with one arm || to the plane of projection,
(ii) a square and its diagonal.
ORTHOGONAL PKOJECTION 119
Ex. 513. Ascertain which of the following relations are unchanged by
projection, (a) generally, (6) in particular cases :
(i) triangle and orthocentre,
(ii) triangle and circumcentre,
(iii) triangle and centroid,
(iv) isosceles triangle,
(v) right-angled triangle,
(vi) parallelogram,
(vii) rectangle,
(viii) rhombus,
(ix) trapezium,
(x) circle,
(xi) a set of equivalent triangles, on the same base and on the same
side of it,
(xii) a set of triangles with the same base and equal vertical angles.
Ex. 514. If the original plane is covered with squared paper, what is
the corresponding pattern on the plane of projection ?
Ex. 515. If a triangle is projected orthogonally, the centroid of the
triangle projects into the centroid of the projection.
The Ellipse.
The most interesting application of the method of orthogonal
projection is that derived from the circle.
The circle projects into an oval curve called an ellipse ; it is
flattened or foreshortened along the lines of steepest slope, while
the dimensions parallel to the plane of projection are unaltered.
If we define the ellipse, for present purposes, as the curve
whose equation is
a'' b^~ '
it is easy to prove that the ellipse is the projection of a circle.
120
OKTHOGONAL PROJECTION
fig. 69.
Let the circle (centre O) be referred to rectangular axes OX,
OY ; OX being i| to the plane of projection.
The coordinates of a point p on the O are 0?i, pn.
Let On = X, pn = Y, radius — a.
Then x' + Y''^ a\
The projections of OX, OY are the perpendicular lines CA, CB;
these shall be the axes for the ellipse.
The coordinates of the point P on the ellipse are ON, PN.
Now ON = O^i = X.
Let PH=y.
Then y = YcosO,
a;2 +
or
V?
cos^^
cos^ 0
But CB, the projection of OY, = a cos B. Let CB = 6.
Then the coordinates of P satisfy the equation
a? h^~
The locus of P is therefore an ellipse whose semiaxes are
CA {a) and CB (6).
The angle properties of the circle do not admit of transference
to the ellipse. But there are many important properties that
ORTHOGONAL PROJECTION 121
may be transferred, and the chief of these are given in the
following exercises.
Ex. 516. Prove the following properties of the ellipse, by
first proving the allied property of the circle, and then carefully
showing that the property admits of projection.
(1) Every chord of the ellipse through C is bisected at C.
(These chords are called diameters.)
(2) The tangents at the extremities of a diameter are parallel.
(3) The locus of the mid-points of a series of parallel chords
is a straight line, namely a diameter.
(4) If a diameter CP bisects chords parallel to a diameter
CD, then CD bisects chords parallel to CP.
(Such diameters are called conjugate.)
(5) The lines joining a point on an ellipse to the extremities
of a diameter are parallel to a pair of conjugate diameters.
(6) A diameter bisects all chords parallel to the tangents at
its extremities.
(7) If a pair of conjugate diameters meet the tangent at
P in T, T', and CD be conjugate to CP, then PT . PT' = CD^.
(8) The chord of contact of the tangents from T is bisected
byCT.
(9) If CT meet the curve in P and the chord of contact of
the tangents from T in N, then
CN . CT = CP2.
(10) Through a point O are drawn two chords ^Op', qOq' ;
and diameters PCP', QCQ' are drawn \\ to the chords. Then
Op . Op : Oq . Oq' - CP^ : CQ^.
(11) Tangents Tp, Tq are drawn from T, and PCP', QCQ'
are the parallel diameters. Then
Tp' : Tg2^Cp2 : CQ^.
122 ORTHOGONAL PROJECTION
(12) PCP', DCD' are a fixed pair of conjugate diameters;
Q is a variable point on the ellipse. QV is drawn || to DC to
meet PCP' in V. Then
QV^ : PV . VP' = CD^ : CP2 = constant.
(13) The area of the ellipse is irah.
(14) A circumscribing parallelogram is formed by the tangents
at the extremities of a pair of conjugate diameters. Its area is
constant and equal to iah.
(15) CP, CD being conjugate semi-diameters,
CP2 + CD^ = constant = a^ -V h\
(16) If all the ordinates of a circle be reduced in a fixed
ratio, the resulting curve is an ellipse.
Ex. 517. By the method of projection, discover some harmonic pro-
perties of the ellipse.
Ex. 518. From a point P on an ellipse a perpendicular PN is drawn to
the major axis ACA'; NQ is drawn parallel to AP and meets CP in Q.
Prove that AQ is parallel to the tangent at P.
CHAPTEH XIII.
CROSS-RATIO.
Definition. A system of points on a straight line is called a
range ; the line is called the base of the range.
Definition. A system of lines through a point is called a
pencil ; the point is called the vertex of the pencil.
Definition. If A, B, C, D be a range of 4 points, and if C, D
be regarded as dividing the line AB (internally or externally),
then -— : — is called a cross-ratio or anharmonic ratio of
CB DB
the range ABCD, and is written {AB, CD}; the sense of lines is
taken into account.
Ranges of equal cross-ratio are called equicross.
A C B D
|< 2"- - -- J< - -1 - -A - 3" - >l
fig. 70.
Ex. 519. Calculate {AB, CD} for the above range. Also calculate
{CD, AB}, {AC, BD}, and all the other cross-ratios obtainable by pair-
ing the points in different ways.
Ex. 520. If a range ABCD is inverted, with respect to a point on the
same line, into a range A'B'C'D', then {AB, CD} = {A'B', CD'} . Examine
what this leads to if A coincides with O, and {OB, CD} is harmonic.
124 CROSS-RATIO
Ex. 521. If a pencil of four lines is cut by two parallel lines in ranges
ABCD, A'B'C'D', then {AB, CD} = {A'B', CD'}.
Ex. 522. If {AB, CD} = {AB, CE}, then the points D and E coincide.
Ex. 523. The projection of a range ABCD on any line is A'B'C'D' ;
prove that {AB, CD} = {A'B', CD'}.
Ex. 524. Investigate the cases
{AB, CD} = 1, {AB,CD}=0, {AB,CD} = oc.
From the definition of cross-ratio, it is clear that one of the
cross-ratios of a harmonic range is equal to — 1.
Ex. 525. If {AB, CD} = {AB, DC}, then ACBD is a harmonic range.
Ex. 526. If A, B, C, D be collinear, and C, D' be the harmonic con-
jugates of C, D respectively with respect to A, B ; then
{AB, CD} = {AB, CD'}.
As four letters admit of twenty-four permutations, the cross-ratio of a
range A, B, C, D can be written down in twenty-four ways. These will not
give rise, however, to twenty-four different cross-ratios.
To begin with, {AB, CD} = {CD, AB} ;
X ,AD ^r^) AC AD AC. DB
for {AB,CD} = ^:^=^3^^,
^ r^r. Ao. CA CB CA.BD AC . DB
and {CD,AB} = ^:3^ = ^^-^ = ^3-^.
In the same way it is shown that
{AB, CD}:={BA, DC}.
Thus {AB, CD} = {CD, AB}:={BA, DC} = {DC, BA},
a group of four equal cross-ratios.
This reduces the possible number of different cross-ratios to six ; and it
will now be shown that these six are generally unequal.
AC . DB
For, let{AB,CD}=^g-^=X.
CROSS-RATIO 125
Then, interchanging the first pair,
^ ' ' CA . BD AC . DB \
Again, {AC, BD}zrl-X.
For AB.CD + AC. DB + AD. BC = 0. (See Ex. 2, p. 4.)
AB . CD AC . DB
•'ad . BC "''AD. BC"*"
„ , ab . cd ab /ad ,^^ „^,
^"*ad7bc--bc/dc=-^a^'^^^
^ AC . DB AC /AD ,.-, __,
^"^a^:bc = -cb/db=-^^^'^^^'
.-. {AC, BD}=1-{AB, CD}
= 1-X.
Interchanging the first pair of {AC, BD},
{CA, BD} = ^-^.
Again, as before,
{BC, AD} = 1-{BA, CD}
_X-1
~ \ *
and {CB, AD}=-^, .
A — 1
We thus have six different cross-ratios,
X 1 1 > 1 ^-1 ^
^' \' ^' 1-X' X ' x-1 •
As a matter of fact, there is seldom any need to consider these various
cross-ratios ; it is customary to use the same cross-ratio throughout a given
calculation, and it does not often become necessary to define which of the
six possible cross-ratios is being used. The comma therefore is generally
omitted, and the cross-ratio written {ABCD}.
Definition. If OA, OB, DC, OD be a pencil of four lines, the
cross-ratio of the pencil is defined to be
sin AOC sin AOD
sin COB * sin DOB'
the sense of angles being taken into account (see p. 5) ; the cross-
ratio of the pencil is written O {AB, CD}.
126 CROSS-RATIO
It is important to notice that the cross-ratio is unaltered if we substitute
for any ray of the pencil (say OB) its prolongation backwards through O
(say OB').
For z COB'= /LCOB + 180° + w.360^
L D0B'= l DOB + 180° + w.360°.
sin C0B'= - sin COB, sin DOB'= -sin DOB.
The cross-ratio is therefore unaltered. In fact, the cross-ratio pertains
to the four complete rays, not to the four lialf-rmjs.
Ex. 627. In Th. 28 it was shown that a system of two lines and the
bisectors of the angles between them is a particular case of a harmonic
pencil. Prove that the cross-ratio of such a pencil, as given by the sine
definition, is equal to - 1.
The cross-ratios of ranges and pencils are brought into relation
by the following fundamental theorem.
Theorem 50.
The cross-ratio of a pencil is equal to the cross-ratio
of the range in which any transversal cuts that pencil.
fig. 71.
To prove that O {AB, CD} = {AB, CD}.
I. As regards sign.
sin AOC , ,, . AC
—. has the same sign as — ;
sniCOB ^ CB'
sin AC D , ,, . AD
— has the same sign as — .
sin DOB ^ DB
.*. O {AB, CD} has the same sign as {AB, CD]
CROSS-RATIO
II. As regards magnitude.
Draw p the perpendicular from O upon A BCD.
A ACC = 1 OA . OC sin AOC,
A COB = 1 DC . OB sin COB,
AAOD = JOA. ODsinAOD,
A DOB = J OD . OB sin DOB.
A AOC AAOD sin AOC sinAOD
127
A COB
A DOB sin COB
sin DOB
-0{AB, CC
^}.
Again,
A AOC = ^p . AC,
ACOB = ijt?.CB,
etc.
, A AOC
AAOD AC
A DOB ~CB
AD
*' A COB '
* DB
= {AB, CD}.
.*. 0{ab, cd} = {ab, cd}.
Theorem 51.
If two lines cut a pencil in the ranges abcd, a'b'c'd',
then {abcd} = {A'B'C'D'}.
fig. 72.
For both {ABCD} and {A'B'C'D'} are equal to O {ABCD}
128
CROSS-RATIO
Ex. 528. Verify graphically the truth of Th. 51.
Ex. 529. Prove that, while the cross-ratios of the ranges ABCD,
A'B'C'D' are equal, the ratios themselves (AB : BC, A'B' : B'C, etc.) are
not equal unless (1) the two lines meet at infinity, or (2) O is at infinity.
Ex. 530. If a transversal be drawn parallel to the ray OD of a pencil
O {ABCD}, and cut the rays OA, OB, OC in P, Q, R respectively, then
PQ: RQ = 0 {AC, BD}.
, Theorem 52.
If two pencils are subtended by the same range, then
the cross-ratios of the pencils are equal.
-AP
\ ^\\/^^ '
y;\ ,z w
fig. 73.
For both P {XYZW} and Q {XYZW} are equal to {XYZW}.
Ex. 531. Verify graphically the truth of Th. 52.
Ex. 532. Examine what becomes of Th. 52, if
(i) P and Q are at infinity,
(ii) XYZW is the line at infinity.
Ex. 533. Show that the two pencils subtended at points P, Q by the
same range XYZW cannot be equiangular unless XYZW is the line at
infinity.
(It may be noted that, if XYZW is the line at infinity, PQXYZW are
concyclic, as a straight line together with the line at infinity is a limiting
form of a circle. But, if PQXYZW are concyclic, z XPY=: z XQY, etc.)
CROSS-RATIO
129
Ex. 534. Consider the pencil D {AYCZ} in fig. 74 :
(i) It is cut by AB ; what range on AB is equicross with {AYCZ}?
(ii) What range on XE is equicross with {AYCZ} ?
fig. 74.
Cross-ratio of a pencil of parallel lines.
If the vertex of a pencil retreats to infinity, the rays become
parallel, and the angles of the pencil become zero. By the
principle of continuity, we may be assured that all transversals
still cut the pencil in equicross ranges ; this property is, however,
obvious from the fact that any two transversals are divided
similarly by a pencil of parallel lines.
The angles of the pencil being zero, it would not appear, at
first sight, that the ordinary definition of the cross-ratio of a
pencil has no application to this case. This difficulty may be
avoided by defining the cross-ratio of a pencil of parallel lines as
the cross-ratio of the range in which any transversal cuts the
pencil.
We may use the property Lt — — = 1 to illustrate the case of a pencil of
parallel lines. For suppose that a circle be drawn with centre O so that
the pencil intercepts arcs AB, BC, CD.
A
"C
fig. 75.
G. S. M. G.
130
CROSS-RATTO
As O retreats towards infinity, let the radius be increased and the angles
be diminished in such a way that the arcs remain finite.
sin AOC /sin AOD / AOC / z AOD
Then
Lt
COB
DOB
;/
zCOB/ z DOB
_ arc AC /arc AD
~ arc CB/ arc^B '
and ultimately the ratios of the arcs become the ratios of the segments of a
transversal line.
Theorem 53.
If {abcd}, {a'b'c'd'} be two equicross ranges, and if
A A', BB', cC be concurrent, then dd' must pass through
the point of concurrence.
fig. 76.
Let O be the point of concurrence of AA', BB', CC'.
If dd' does not pass through O, let OD cut A'B' in D".
Then {A'B', C'D"} = {AB, CD}
- {A'B', C'D'}.
A'C' . D"B' _ A'C' . D'B'
C'B' . A'D" ~ C'B' . A'D' '
D"B' _ D'B'
*• a^'~a'd"
.*. D" coincides with D',
.', DP' passes through O.
CROSS-RATIO 131
Note. This theorem and Theorem 51 could be stated as
theorem and converse. It must be carefully noted that it is
generally not true that, if {ABCD} ^ {A'B'C'D'}, then AA', BB', CC',
DD' are concurrent.
Ex. 535. Examine the particular case in which {ABCD}, {A'B'C'D'}
are similar.
Ex. 536. Place two similar ranges {ABCD}, {A'B'C'D'} in such a
position that AA', BB', CC, DD' are not concurrent.
Theorem 54.
If two equicross ranges {pxyz}, {px'y'z'} have a point
p in common, then xx', yy', zz' are concurrent.
This is a particular case of Theorem 53.
Ex. 537. Prove this theorem without assuming Th. 53.
9—2
132
CROSS-RATIO
Theorem 55.
If p {XYZW}, Q {XYZW} be two equicross pencils, and if
X, Y, z be coUinear, then w is on the line XYZ.
'. 78.
If W does not lie on XYZ, let PW, QW cut XYZ in A, B
pectively.
Then
{XY, ZA} = P {XY, ZW}
= Q{XY, ZW}
= {XY, ZB}.
. XZ . AY XZ . BY
* ZY.XA "ZY.XB'
AY BY
•• XA~XB'
.*. A and B coincide,
.-. W lies on XYZ.
CROSS-RATIO 133
Theorem 56.
If two equicross pencils p {abcd}, q {abcd} have a ray
PQA in common, then BCD are collinear.
fig. 79.
This is a particular case of Theorem 55.
Ex. 538. Prove Th. 56 without assuming Th. 55.
Ex. 539. Prove that in fig. 77 the intersections of XY', X'Y ; of XZ',
X'Z ; of YZ', Y'Z He on a hne through P. (Consider two of the above
points.)
Ex. 540. Join the intersection of QB, PC to that of QC, PB ; that of
QB, PD to that of QD, PB ; that of QC, PD to that of QD, PC. Prove that
these three Unes meet on PQ.
Cross-ratios and Projection.
We have seen that angle properties as a rule are destroyed
by orthogonal projection. One important set of angle relations,
hovs^ever, are undisturbed ; namely, those connected with cross-
ratios. The reader will be able to appreciate the importance of
cross-ratio in view of the following theorems.
A range of points is equicross with the range obtained
by projecting these points.
A pencil of lines is equicross with the pencil obtained
by projecting these lines.
The proofs are left to the reader.
It follows from the above theorems that harmonic properties
of points and lines are unaltered by projection.
134 CROSS-RATIO
Exercises on Chapter XIII.
Ex. 541. Find a point on a given line such that if it be joined to three
given points in a plane with the line, any parallel to the line is divided in a
given ratio by the three joins.
Ex. 542. Four fixed points on a circle subtend at a variable
point on the circle a pencil of constant cross-ratio.
Ex. 543. Four fixed tangents to a circle meet a variable tangent
to the circle in a range of constant cross-ratio.
(Consider the pencil subtended at the centre.)
Ex. 544. If four points are collinear, their polars vritb. respect
to a circle are concurrent; the cross-ratio of the pencil so formed
is equal to that of the range formed by the four points.
Ex. 545. X is the vertex of a fixed angle; PAB is a transversal which
turns about a fixed point P and cuts the arms of the angle in A, B ; O, O'
are two fixed points collinear with X. OA, O'B meet in Q. Prove that the
locus of Q is a straight line.
(Consider a pencil formed by PX and three positions of the transversals
PAiBi, PA2B2, PA3B3.)
Ex. 546. With the notation of the preceding exercise, let O, O' be col-
linear with P instead of X. Prove that the locus of Q is a straight line
through X.
(Consider a pencil formed by POO' and three positions of the trans-
versal.)
Ex. 547. Prove that if the sides of the triangle O1O2O3 pass through
the vertices of the triangle U1U2U3, and Ai be any point on U2U3, and O3A1
meet Ui U3 in A2, and O2A1 meet U1U2 in A3, then Oi, A2, A3 are collinear.
(Consider pencils whose vertices are Ai , U^.)
Ex. 548. Three points F, G, H are taken on the side BO of a triangle
ABC ; through G any line is drawn cutting AB and AC in L and M re-
spectively; FL and HM intersect in K; prove that K lies on a fixed straight
line passing through A.
Ex. 549. The three sides of a varying triangle ABC pass each through
one of three fixed collinear points P, Q, R. Further, A and B move along
fixed lines; show that C also moves on a fixed line, concurrent with the
other two.
CROSS-RATIO 135
Ex. 550. A straight line drawn through a point P meets two fixed
straight lines in the points L and M. The straight lines joining L and M
to a point Q meet the fixed straight lines again in the points M' and L'.
Show that if P and Q are fixed, L'M' passes through a fixed point.
Ex. 551. Show that the lines joining the centres of the escribed circles
of a triangle to the corresponding vertices of the pedal triangle are con-
current.
Ex. 552. Prove that the lines joining the centres of the escribed circles
of a triangle to the middle points of the corresponding sides are concurrent.
Ex. 553. A', B', C are the mid-points of the sides of the triangle ABC,
and any line is drawn to meet the sides of the triangle A'B'C in K, L, M.
AK, BL, CM meet the sides of ABC in K', L', M' respectively. Prove that
K'L'M' is a straight line.
Ex. 554. If A', B', C be three points on the sides of a triangle ABC
such that AB' . BC . CA' = AC . BA' . CB' and X, Y, Z be the mid-points of
B'C C'A', A'B', then AX, BY, CZ are concurrent.
Ex. 555. Two points X, Y separate harmonically each of the three pairs
of points P and P', Q and Q', R and R'. Prove that
{PP'QR} = {P'PQ'R'}.
CHAPTEE XIV.
THE PRINCIPLE OF DUALITY.
THE COMPLETE QUADRILATERAL AND
QUADRANGLE.
The reader may have noticed that there exists in plane
geometry a certain duality, by which many properties of points
have, as their counterpart, corresponding properties of lines.
For instance : —
2 points define 1 line. 2 lines define 1 point.
3 points define 3 lines. 3 lines define 3 points.
4 points define 6 lines. 4 lines define 6 points.
etc. etc.
A point moving under cer- A line moving under cer-
tain conditions defines a curve, tain conditions defines a curve,
the locus. the envelope.
If a point lies in a fixed If a straight line passes
line, its polar with respect to a through a fixed point, its pole
circle passes through a fixed with respect to a circle lies in
point. a fixed line.
This duality has obvious limitations, though a more extended
study of geometry will show that it reaches further than would
appear at first sight : e.g. there would at first sight seem to be
no point-system corresponding with a line-system of two lines at
right angles.
PRINCIPLE OF DUALITY
137
However, there are many cases of duality that may be cited
at this stage.
In order to exhibit the matter in the most striking way, it is
convenient to use two new terms : —
Definition. The join of Definition. The meet of
two points is the unlimited two lines is the point defined
line defined by the two points. by the two lines (by their
intersection).
It is also convenient to denote points by large letters, and
lines by small letters : AB is the join of points A, B ; ah is the
meet of lines a, h.
Using this notation : —
I
C/
■rp
fig. 80.
A range of four fixed points
A, B, C, D together with a
varying point P define a pencil
of constant cross-ratio.
If two equicross ranges
A BCD, A'B'C'D' be placed so
that the lines A A', BB', CC' are
concurrent, then DD' will be
concurrent with these three
lines.
A pencil of four fixed lines
a, b, c, d together with a vary-
ing line p define a range of
constant cross-ratio.
If two equicross pencils
abcd^ a'h'c'd' be placed so that
the points aa\ bb\ cc' are col-
linear, then dd' will be collinear
with these three points.
138
COMPLETE QUADRILATERAL
An interesting case is that of the complete figures defined by
four lines and four points.
Definitions.
Four lines together with Four points together with
their six meets form a com- their six joins form a com-
plete quadrilateral (or four- plete quadrangle (or four-
line).
point).
fig. 83.
fig. 82.
The four lines AB, BC, CD,
DA are called sides.
The meet of any two sides
is called a vertex ; the vertices
are the six points A, B, C, D,
E, F.
Opposite vertices are ver-
tices that do not lie on the
same side (A, C ; B, D ; E, F).
The join of two opposite
vertices is called a diagonal ;
these are three in number, AC,
BD, EF.
The four points ah, be, cd,
da are called vertices.
The join of any two vertices
is called a side; the sides are
the six lines a, b, c, d, e, f.
Opposite sides are sides
that do not pass through the
same vertex (a, c; b, d\ e,f).
The meet of two opposite
sides is called a diagonal-
point; these are three in
number, ac, bd, ef.
COMPLETE QUADRILATERAL
139
We will now prove the important harmonic property of the
, , fquadrilateral
complete i ^ , , .
^ (.quadrangle
Before proving this, it should be noted that
If ABCD is a range of points
and P a point not lying on the
same line, P {ABCD} signifies
the cross-ratio of the pencil
PA, PB, PC, PD,
If abed is a pencil of lines
and J) a line not passing through
the same point, p {abed} signifies
the cross -ratio of
pa, pb, pe, pd.
the range
Theorem 57.
In a complete quadri-
lateral, on each diagonal
there is a harmonic range
formed by its meets with
the other two diagonals to-
gether with two vertices of
the quadrilateral.
To prove {EF, PQ} a har-
Theorem 58.
In a complete quad-
rangle, through each dia-
gonal-point there is a har-
monic pencil formed by its
joins to the other two dia-
gonal-points together with
two sides of the quadrangle.
To prove {ef, pq] a harmonic
{EF, PQ}=B{EF, PQ}
= {CA, RQ}
= D {CA, RQ}
^{FE, PQ}^.
Since the points E, F have
been interchanged without
altering the value of the cross-
ratio, {EF, PQ} is harmonic.
{^/, P9] = ^ {e/, pq}
= {c«, rq}
= d{ca, rq}
= {A nV
Since the lines e, f have
been interchanged without
altering the value of the cross-
ratio, [efj pq} is harmonic.
* This method of proof may be remembered as follows : the range on
diagonal 1 is projected on to diagonal 2, and back again on to diagonal 1;
using the two vertices that lie in diagonal 3.
140
COMPLETE QUADRILATERAL
The above proof is of interest as bringing out the principle
of duality. The following proof, however, may be preferred for
ordinary purposes.
fig. 84.
Fig. 84 represents a com-
plete quadrilateral.
To prove that {TU, XY} is a
harmonic range.
Fig. 85 represents a com-
plete quadrangle.
To prove that Z {QR, TUj
is a harmonic pencil.
Consider the triangle STU.
Since SX, TR, UP are concurrent,
TX UR SP
UX * SR ' f P ~ ~
Again, since P, R, Y are collinear,
TY UR SP
U Y ■ SR ' TP
1.
[Ce
\^Menelaus.
TX
UX
TY
UY
.*. {TU, XY} is a harmonic range.
Hence Z {TU, QR} is a harmonic pencil.
COMPLETE QUADRILATERAL 141
Ex, 556. Prove the above theorem for the other two diagonals of the
quadrilateral, and for the other two diagonal points of the quadrangle.
Ex. 557. AB is parallel to DC ; AC, BD meet in Q ; DA, CB in P. Prove
that PQ bisects AB and DC.
Ruler construction for the fourth point of a harmonic
range.
't
/
m
'<
,Q\»
^^
' /
\ "^
^^,^
^ ^ ^
: /
^
^ ^
X,
/
^^
^~
N.
8^
fig.
C
86.
Given three coUinear points A, B, C ; to find the point D such
that {AC, BD} shall be harmonic.
Through A draw any two lines AP, AQ.
Through B draw any line BQP cutting the two former lines
in Q and P respectively.
Join CQ, CP. Let these joins cut AP, AQ in R, S respectively.
Join RS and produce it to meet ABC in D.
Then, by the harmonic property of the quadrilateral FRQS,
{AC, BD} is harmonic.
Note. This ruler construction for a fourth harmonic point
is important, as showing that the idea of a harmonic range can
be developed without any reference to measurements of lines or
angle; in other words, can be put on a non-metrical or pro-
jective basis.
142 SELF-POLAR TRIANGLE
Ex. 558. Perform the above construction for the point D, satisfying
yourself that the same point is obtained however the lines AP, AQ, BQP are
varied.
Ex. 559. Bisect a line AC by the above method. [D will be at in-
finity.]
Ex. 560. Show that, if one diagonal of a complete quadrilateral is
parallel to the third (the exterior) diagonal, then the second diagonal bisects
the third.
Ex. 561. Apply the harmonic property of the quadrilateral to the case
of the parallelogram, considering all three diagonals.
Self-polar Triangle. >
The reader is reminded of the following theorems proved in
Chapter vii.
Th. 31. If a straight line is drawn through any point
to cut a circle, the line is divided harmonically by the
circle, the point, and the polar of the point with respect
to the circle.
Th. 32. If the polar of a point P with respect to a
circle passes through a point Q, then the polar of Q
passes through p.
Th. 33. Two tangents are drawn to a circle from a
point A on the polar of a point B ; a harmonic pencil is
formed by the two tangents from A, the polar of B and
the line ab.
Ex. 562. Let the polars of poinifs A, B, C form a triangle PQR. Prove
that the polars of the points P, Q, R are the sides of the triangle ABC.
Ex. 563. Draw the polar of a point A. On this polar take a point B.
Draw the polar of B, passing through A (why ?) and cutting the polar of A
in C. Prove that AB is the polar of C ; i.e. that eacli side of the txiangle
ABC is the polar of the opposite vertex.
SELF-POLAR TRIANGLE 143
Definition. If a triangle be such that each side is the polar
of the opposite vertex with respect to a given circle, the triangle
is said to be self-polar or self-conjugate with respect to the
circle; and the circle is said to be polar with respect to the
triangle.
From Ex. 563 it is seen that an infinite number of triangles may be
drawn self-polar with respect to a given circle. One vertex may be taken
anywhere in the plane ; the second is then limited to a certain line ; and
when the second is fixed, the third is thereby fixed.
On the other hand, it will appear from Exs. 564, 565, that a given triangle
has only one polar circle.
Ex. 564. The centre of a circle, polar with respect to a given triangle,
is the orthocentre of the triangle.
Ex. 565. If H be the orthocentre of a ABC, and AD, BE, CF the alti-
tudes, then
HA . HD = (rad. of polar circle) 2,
and similarly
HB . HEzr:(rad. of polar circle) 2 =HC . HF,
the sense, of lines being taken into account,
Ex. 566. A triangle self-polar with respect to a real circle cannot be
acute-angled.
Ex. 567. What is the polar circle of a right-angled triangle?
Ex. 568. An isosceles triangle ABC has base 2a and vertical angle (A)
120°. Show that the radius of the polar circle is a ^^2. If the polar circle
cuts AC in P, show that z ABP = 15°.
Ex. 569. Tlie sides of a triangle are divided harmonically by its
polar circle.
Ex. 570. A triangle self-polar with respect to a point-circle is right-
Ex. 571. What does a self-polar triangle become if one vertex coincides
with the centre of the circle ?
Ex. 572. If a circle consists of a straight line and the line at infinity,
what do its self -polar triangles become ?
144 SELF-POLAR TRIANGLE
Ex. 573. The angle A of a triangle ABC is obtuse ; AD, BE, CF are the
altitudes ; H the orthocentre. The polar circle cuts AC in P and Q. Show
that EP2= EA . EC, and that H, F, P, D, B, Q are concyclic.
Ex. 574. If circles are described in the sides of a triangle as diameters,
they are cut orthogonally by the polar circle of the triangle.
Theorem 59.
If a quadrangle be inscribed in a circle, the triangle
formed by the diagonal points is self-polar with respect
to the circle.
fig. 87.
We will prove that the side TU of the triangle TUZ is the
polar of the vertex Z.
By Theorem 58 T {ZU, SQ} is a harmonic pencil.
.'. the pencil is cut by SQ in the harmonic range {ZX, SQ}.
.*. X is on the polar of Z. Th. 31.
Again, T {ZU, SQ} is cut by PR in the harmonic range
{ZY, PR}.
.-. Y is on the polar of Z. Th. 31.
.*. XY or TU is the polar of Z.
Similarly it may be shown that UZ is the polar of T and ZT
the polar of U.
Ex. 575. Prove in detail that UZ is the polar of T, and ZT the polar
of U.
SELF-POLAR TRIANGLE 145
Theorem 60.
If a quadrilateral be circumscribed about a circle, the
triangle formed by the diagonals is self-polar with respect
to the circle.
fig. 88.
We will prove that the vertex Z of the triangle XYZ is the
pole of XY.
By Theorem 57, {XZ, QS} is a harmonic range.
.'. U {XZ, QS} is a harmonic pencil.
.*. UZ passes through the pole of UX. Th. 33.
Again, T {XZ, QS} is a harmonic pencil.
.*. TZ passes through the pole of TX. Th. 33.
.*. Z is the pole of XY.
Similarly it may be shown that X is the pole of YZ, and Y
the pole of ZX.
Ex. 676. Prove in detail that X is the pole of YZ, and Y the pole
of ZX.
G. S. M. G. 10
146 TRIANGLES IN PERSPECTIVE
Triangles in perspective.
Definition. Two figures are said to be in perspective if the
joins of corresponding pairs of points are all concurrent.
Theorem 61.
(Desargues' Theorem*.)
If two triangles are such that the lines joining their
vertices in pairs are concurrent, then the intersections
of corresponding sides are coUinear.
C
fig. 89.
The triangles ABC, A'B'C' are such that AA', BB', CC' meet
at O.
Let BC, B'C' meet at P; CA, C'A' at Q ; AB, A'B' at R. Let
OAA' cut BC in S, B'C' in S'.
* Gerard Desargues (born at Lyons, 1593 ; digd, 1662).
TRIANGLES IN PERSPECTIVE 147
To 'prove that PQR is a straight line.
{PBSC} ^ {PB'S'C'} as both ranges lie on the pencil O {PBSC}.
.-. A {PBSC} = A' {PB'S'C},
i.e. A{PROQ}-A'{PROQ}.
These two equicross pencils, therefore, have a line OAA' in
common.
.'. P, Q, R are collinear. Th. 56.
Definition. The point O is called the centre of perspective,
and the line PQR the axis of perspective of the two triangles
ABC, A'B'C' in fig. 89.
Ex. 577. Prove Th. 61 by considering equicross pencils with vertices
at B and B' (instead of A and A').
Ex. 678. Investigate whether Th. 61 can be extended to the case of
polygons in perspective.
Ex. 579. Prove Th. 61 for the case in which the triangles ABC, A'B'C
are not in the same plane. Hence prove the theorem for coplanar triangles
by rotating the line OAA' about O till it comes into the plane OBB'CC.
Ex. 580. Prove Th. 61 by considering fig. 89 as the representation in
piano of three planes meeting at O and cut by the planes ABC, A'B'C.
Ex. 681. Prove the converse of Th. 61.
Ex. 682. Prove that triangles that are similar and similarly situated
(i.e. sides parallel) are in perspective. Where is the axis of perspective ?
Ex. 583. Investigate whether Ex. 582 can be extended to polygons.
Ex. 684. Consider the case of triangles that are congruent and
similarly situated.
10—2
148 PRINCIPLE OF DUALITY
Note on Three-dimensional Geometry.
The dual relation of point and line is confined to two-
dimensional geometry.
In three dimensions, the point corresponds to the plane, the
line occupying an intermediate position.
Thus:
Two points determine a line. Two planes determine a line.
Three points determine a Three planes determine a
plane, unless they are all on point, unless they all contain
the same line. the same line.
Two lines, in the same Two lines, through the same
plane, determine a point. point, determine a plane.
A point and a line deter- A plane and a line deter-
mine a plane, unless the line mine a point, unless the line
passes through the point. --^ lies in the plane,
etc. etc.
Again, consider the five regular solids. They may be grouped
as follows :
Tetrahedron (3 corners, 6 edges, 3 faces).
Cube (8C, 12E, 6f). Octahedron (8F, 12E, 6c).
Dodecahedron (20C, 30E, 12f). Icosahedron (20F, 30E, 12C).
The point-plane correspondence appears very clearly when
we take stock of the cross-ratio properties of three dimensions.
We should begin with the definitions of :
(1) Cross-ratio of four points on the same line (a range of
points).
(2) Cross-ratio of four planes containing the same line (a
sheaf of planes), this being defined by means of the angles between
the planes.
PRINCIPLE OF DUALITY 149
In addition there would be the definition of the
(3) Cross-ratio of four lines, in a plane, through a point (a
pencil of lines).
There would then follow a number of theorems such as the
following :
The joins of a point to the . The intersections of a plane
four points of a range give a with the four planes of a sheaf
pencil equicross with the range. give a pencil equicross with the
sheaf.
The planes determined by The points determined by
a line and the four points of a a line and the four planes of a
range give a sheaf equicross sheaf give a range equicross
with the range. with the sheaf.
The proofs of the theorems may be left to the reader, who
will find that these principles admit of further development*.
Exercises on Chapter XIY.
Ex. 585. A straight line meets the sides BC, CA, AB of a triangle in
the points P, Q, R respectively; BQ and CR meet at X and AX meets BC
at P'. Show that P and P' are harmonic conjugates with respect to B and C.
If X is the orthocentre of ABC, show that XP is perpendicular to the
straight line joining A to the middle point of BC.
Ex, 586. The collinear points ADC are given; CE is any other fixed
line through C, E is a fixed point and B is any moving point on CE. The
lines AE, BD intersect in Q; the lines CQ, DE in R ; and the lines BR, AC
in P. Prove that P is a fixed point as B moves along CE.
Ex. 587. If a line drawn through the intersection O of the diagonals of
a quadrilateral cuts one pair of opposite sides in P, P', so that OP=P'0,
and cuts the other pair in Q, Q', show that PQ=Q'P'.
* See Reye's Geometry of Position, translated by Holgate (the Macmillan
Company).
150 NOTE ON THREE-DIMENSIONAL GEOMETRY
Ex. 588. Perpendiculars at B, C to the sides BA, CA of a triangle ABC
meet the opposite sides in P, Gt ; and the tangents to the circumcircle at B,
C meet in R. Prove that P, Q, R are collinear.
Ex. 589. A quadrilateral is such that pairs of opposite sides have the
same sum. If O be the orthocentre of the triangle formed by the diagonals,
then O is also the in-centre of the quadrilateral.
Ex. 590. Two tangents to a circle are fixed; two others are drawn so as
to form with the two fixed tangents a quadrilateral having two opposite
sides along the fixed tangents ; show that the locus of the intersection of
internal diagonals of this quadrilateral is a straight line, and find its
position.
Ex. 591. ABCD is a quadrilateral inscribed in a circle whose centre is
O; AB, CD intersect in E; AD, BC intersect in F; AC, BD intersect in G.
Prove that OG is perpendicular to EF; and that BC, AD subtend equal
angles at the foot of the perpendicular from O upon EF.
Ex. 592. Prove that the circle on each of the diagonals of a quadri-
lateral as diameter is orthogonal to the polar circle of each of the four
triangles formed by the sides of the quadrilateral.
Ex. 593. Prove that the midpoints of the diagonals of a complete
quadrilateral are collinear.
(Let ABCDEF be the quadrilateral; EF being the third diagonal. Let
P, Q, R be the mid-points of AC, BD, EF. Prove that as PQE, PQF are
each I of the quadrangle ABCD, )
Ex. 594. ABC, A'B'C, A"B"C", are three triangles in perspective, and
BC, B'C, B"C" are parallel. Prove that the line joining the intersections
of AB, A'B', and AC, A'C, is parallel to the line joining the intersections of
A'B', A"B", and A'C, A"C".
Ex. 595. The lines EF, FD, DE which join the points of contact D,
E, F of the inscribed circle of a triangle with the sides cut the opposite sides
X, Y, Z. Prove that the mid-points of DX, EY, FZ are collinear.
Ex. 596. Show that in a complete quadrangle the three sides of the
harmonic triangle are met by the sides of the quadrangle in 6 points, other
than the vertices of the harmonic triangle, which lie by threes on four
straight lines.
MISCELLANEOUS EXERCISES.
Ex. 597. ABC is a triangle ; D, E, F are the feet of the perpendiculars.
Prove that, if the triangles FBD, EDC are equal in area, AB is equal to AC.
Ex. 598. In a given circle show how to inscribe a triangle ABC such
that the angle ABC is given and the sides AB, AC pass through given
points.
Ex. 599. From a fixed point A straight lines ABC, AEF are drawn to
meet two fixed lines in B, C and E, F. Prove that the circles circumscribing
the triangles ABE, ACF intersect at a constant angle.
Ex. 600. The perpendiculars drawn to the sides of a triangle at the
points in which they are touched by the escribed circles are concurrent.
Ex. 601. Three circles have two common points O and C, and any
straight line through O cuts them in points P, Q, and R. Prove that the
circumscribing circle of the triangle formed by the tangents at P, Q, R passes
through O'.
Ex. 602. Draw a straight line from the vertex A of a triangle ABC
meeting BC in P so that AP2=BP. CP, considering the cases in which P
(i) is, (ii) is not, situated between B and C.
Ex. 603. A, B, C, D are four points in a plane: points P, Q, R are
taken in AD, BD, CD respectively such that
AP : PD = BQ : QDzzzCR : RD .
Show that the three lines joining P, Q, R to the middle points of BC, CA,
AB respectively are concurrent.
Ex. 604. Prove that the locus of the middle points of the sides of all
triangles which have a given orthocentre and are inscribed in a given circle
is another circle.
Ex. 606. A straight line drawn parallel to the median AD of an isosceles
triangle ABC whose angle A is a right angle cuts the sides AB, AC in P and
Q. Show that the locus of M, the intersection of BQ, CP, is a circle; and
that, if N is the middle point of PQ, M N touches this circle.
Ex. 606. A straight line drawn through the vertex of a triangle ABC
meets the lines DE, DF, which join the middle point D of the base to the
middle points E, F of the sides, in X, Y ; show that BY is parallel to CX.
152 MISCELLANEOUS EXERCISES
Ex. 607. The points of contact of the escribed circles with the sides
BC, CA, AB produced when necessary, are respectively denoted by the letters
D, E, F with suffixes 1, 2 or 3 according as they belong to the escribed circle
opposite A, B or C. BEg, CFg intersect at P; BEj, CF^ at Q; EgFg and
BC at X ; FgDj and CA at Y ; D^ E^ and AB at Z. Prove that the groups
of points A, P, Dj, Q ; and X, Y, Z ; are respectively collinear.
Ex. 608. The opposite sides of the hexagon ABCDEF are parallel, and
the diagonal CF is parallel to the sides AB and DE; BC, AF intersect in
P, CD, EF in Q, and BD, AE in R; show that P, Q, R are in one straight
line.
Ex. 609. Show that, if O be any point on the circumcircle of the
triangle ABC, and OL be drawn parallel to BC to meet the circumcircle in
L, then will LA be perpendicular to the pedal line of O with respect to the
triangle.
Ex. 610. ABC is a triangle inscribed in a circle, and tangents to the
circle at A, B, C cut BC, CA, AB respectively in the points A', B^^C.
Show that the middle points of A A', BB', CC lie on the radical axis of
the circumcircle and nine-points circle.
Ex. 611. If ABC is a triangle and DEF its pedal triangle, the perpen-
diculars from A, B, C upon EF, FD, DE respectively are concurrent.
Ex. 612. ABC is a triangle right-angled at C. The bisector of the angle
A meets BC in D, the circumcircle in G, and the perpendicular to AB
through the circumcentre in F. Prove that 2FG = AD. Hence (or other-
wise) construct a right-angled triangle, given the hypotenuse and the length
of the line drawn bisecting one of the acute angles and terminated by the
opposite side.
Ex. 613. A point O is taken within an equilateral triangle ABC such
that the angles AOB, BOC, CCA are in the ratios 3:4:5. AD is drawn
perpendicular to BC, and CD is joined. Show that each of the triangles
into which ADC is divided by OA, CD, OC is similar to one of the triangles
into which ABC is divided by OA, OB, OC.
Ex. 614. Two circles intersect in the points B, D ; a straight line ABC
cuts the circles in A, C; AD, CD cut the circles again in P, Q ; AQ, CP
meet in R ; prove that DPQR is a cyclic quadrilateral.
Ex. 615. If S, S' are the centres of similitude of two circles, prove that
the circles subtend equal angles at any point on the circle whose diameter is
SS'.
Ex. 616. A circle S passes through the centre of another circle S';
show that their common tangents touch S in points lying on a tangent to S'.
MISCELLANEOUS EXERCISES 153
Ex. 617. Three circles have two common points O and O', and any
straight Hne through O cuts them in points P, Q, R. Prove that the circum-
scribing circle of the triangle formed by the tangents at P, Q, R passes
through O'.
Ex. 618. A quadrilateral A BCD is inscribed in a circle, and through a
point E on AB produced a straight line EFG is drawn parallel to CD and
cutting CB, DA produced in F, G respectively. Show how to draw the
circle that passes through F and G and touches the given circle.
Ex. 619. In a triangle AiBjC^ a circle is inscribed, touching the sides in
A2B2C2 ; and so on. Find the values of the angles of the triangle A^B„C„, and
give a construction for the directions of the sides when n is made infinite.
Ex. 620. The lines WAX, XBY, YCZ, ZDW bisect the exterior angles
of the convex quadrilateral ABCD. Show that an infinite number of quadri-
laterals can be inscribed in XYZ W whose sides are parallel respectively to the
sides of ABCD, and whose perimeters are equal to the perimeter of ABCD.
Ex. 621. A, B, C are three given points. Show how to describe a square
having one vertex at A so that the sides opposite to A shall pass through B, C
respectively.
Ex. 622. Any point P is taken on the base BC of a triangle ABC, and a
line PL parallel to BA meets AC in L, while a line PM parallel to CA meets
AB in M. Show that the triangle PLM is a mean proportional between the
triangles BMP, PLC.
Ex. 623. GAB is a triangle. Any circle through A, B meets OA at P
and OB at Q. PQ meets AB at X, PB meets AQ at Y. Find the locus of Y,
and show that XY passes through a fixed point.
Ex. 624. Prove that the radical axes of a fixed circle and the several
circles of a coaxal system meet in a point. State the theorems which may
be obtained by inverting this theorem with respect to (i) a limiting point,
(ii) a point of intersection of the coaxal circles, (iii) any other point in the
plane.
Ex. 625. A trapezium ABCD has the opposite sides AB, CD parallel.
Shew that the common chord of the circles described on the diagonals AC,
BD as diameters is perpendicular to AB and CD, and concurrent with AD
and BC.
Ex. 626. Given three points A, B, C on a circle, determine geometrically
a fourth point D on the circle, such that the rays PC, PD may be harmonic
conjugates with respect to the rays PA, PB, where P is any point in the circle.
Show further that the intersection of AC, BD, that of AB, CD, that of
the tangents at A and D, and that of the tangents at B and C are collinear.
ID— 5
154 MISCELLANEOUS EXERCISES
Ex. 627. Find the locus of the centre of a circle which bisects the
circumferences of two given circlefe.
Ex. 628. O is the radical centre of three circles. Points A, B, C are
taken on the radical axes and AB, BC, CA are drawn. Prove that the six
points in which these meet the three given circles lie on a circle.
If radii vectores are drawn from O to these six points they meet the
three given circles in six points on a circle and its common chords with the
three circles meet in pairs on OA, OB, OC.
Ex. 629. On a given chord AB of a circle a fixed point C is taken, and
another chord EF is drawn so that the lines AF, BE and the line joining C to
the middle point of EF meet in a point O ; show that the locus of O is a
circle.
Ex. 630. If O be the centroid of the n points A, B, C,.,. and if P be
any variable point, then A P^ + B P'-^ + C P^^ + . . . = ?i . O P- + constant.
If ABC... be a regular polygon inscribed in a circle, O the centre, and P
any point on the circumference of this circle, then the centroid of the feet of
the perpendiculars from P on OA, OB, OC, ...will lie on a fixed circle.
Ex. 631. If A, B, C are three collinear points and P is any point what-
ever, prove that BC . PA^ + CA . PB-' + AB . PC^^ - BC . CA . AB. Find the
radius of the circle which touches the circles described on AB, BC, AC as
diameters.
Ex. 632. Prove that the tangents to the circumcircle of the triangle
ABC at the vertices meet the opposite sides in collinear points.
Ex. 633. If L, L' are the limiting points of a family of coaxal circles,
prove that any circle through L, L' cuts the family orthogonally, and that if
PP' is a diameter of this circle, then the polars of P with respect to the
family pass through P'.
Ex. 634. A line drawn through L, a limiting point of a coaxal system
of circles, cuts one of the circles at A and B. The tangents at A and B cut
another circle of the system at P, Q and R, S respectively. Shew that PR
and QS subtend equal angles at L.
Ex. 635. P, Q are any two points; PM is drawn perpendicular to
the polar of Q with respect to a circle, and QN is drawn perpendicular
to the polar of P ; if O is the centre of the circle, prove that
PM:QN = OP:OQ.
MISCELLANEOUS EXERCISES 155
Ex. 636. If P be the extremity of the diameter CP of any circle through
L, C, where L, L', C, C are the limiting points and centres of two fixed
circles and L lies within the circle with C as centre, then the polar of P with
regard to the circle with C as centre passes through a fixed point.
Ex. 637. A chord of a fixed circle is such that the sum of the squares
of the tangents from its extremities to another fixed circle is constant ;
prove that the locus of its middle point is a straight line.
Ex. 638. A circle touches two given circles in P and P', and intersects
their radical axis in Q and Q'. Prove that PP' passes through one of the
centres of similitude of the given circles, and that the tangents at Q and Q'
are parallel to a pair of common tangents of the given circles.
Ex. 639. State (without proof) the chief properties of any geometrical
figure which persist after inversion. If Gl, Q' are inverse points with respect
to a circle B, and R, R' are the inverse points of Q, Q' with respect to an
orthogonal circle C, prove that R, R' are inverse points with respect to the
circle B.
Ex. 640. Two circles intersect in A and B, and a variable point P on
one circle is joined to A and B, and the joining lines, produced if necessary,
meet the second circle in Q and R. Prove that the locus of the centre of
the circle circumscribing PQR is a circle.
Ex. 641. Two squares have a common angular point at A and their
angular points taken in order the same way round are respectively A, B, C, D
and A, B', C, D'. Prove that the lines BB', CC, and DD' are concurrent.
Ex. 642. A, B, A', B' are given points, and PQ is a given straight line.
Find points C, C in PQ such that the area of the triangles ABC, A'B'C
shall be equal, and CC shall be of a given length.
Ex. 643. The middle points of the sides of a plane polygon A are joined
in order so as to form a second polygon B ; prove that about this polygon B
either an infinite number of polygons other than A, or no other can be cir-
cumscribed with their sides bisected at the corners of B, according as the
number of sides is even or odd.
Ex. 644. A circle is inscribed in a triangle ABC touching the sides at
P, Q, R. Show that the diameter of the circle through P, the line QR, and
the line joining A to the middle point of BC, are concurrent.
156 MISCELLANEOUS EXERCISES
Ex. 645. A common tangent touches two circles in P and Q re-
spectively ; show that P and Q are conjugate points * with regard to any
coaxal circle.
Ex. 646. If one pair of opposite vertices of a square is a pair of con-
jugate points with respect to a circle, so will be the other pair.
Ex. 647. Having given two non-intersecting circles; draw the longest
and the shortest straight line from one to the other, parallel to a given
straight line.
Ex. 648. POP', QOQ' are two chords of a fixed circle and O is a fixed
point. Prove that the locus of the other intersection of the circles POQ,
P'OQ' is a second fixed circle.
Ex. 649. The points Q and R lie on the straight line AC and the
point V on the straight line AD ; VQ meets the straight line AB in Z, and
VR meets AB in Y: X is another point on AB : XQ meets AD in U, and XR
meets AD in W. Prove that YU, ZW, AC are concurrent.
Ex. 650. The opposite sides of the hexagon ABCDEF are parallel, and
the diagonal CF is parallel to the sides AB and DE ; BC, AF intersect in P,
CD, EF in Q, and BD, AE in R ; show that P, Q, R are in one straight
line.
Ex. 661. A, B are two fixed points, and a variable circl| through them
cuts a fixed circle in C, D. Prove that the line joining the intersections of
AC, BD and AD, BC passes through a fixed point.
Ex. 652. Having given six points A, B, C, A', B', C such that A'B is
parallel to AB', B'C is parallel to BC, and C'A is parallel to CA', prove that
if A' B'C are collinear, ABC also are collinear.
Ex. 653. The angles APB, AQB subtended at two variable points P, Q
by two fixed points A, B differ by a constant angle, and the two ratios AP/BP,
AQ/BQ are proportionals. Show that if P describes a circle, Q describes
either a circle or a straight line.
Ex. 654. A straight line drawn through the vertex of a triangle ABC
meets the lines DE, DF, which join the middle point D of the base to the
middle points E, F of the sides in X, Y ; show that BY is parallel to CX.
* Two points are said to be conjugate with respect to a circle if the
polar of each point passes through the other.
MISCELLANEOUS EXERCISES 157
Ex. 655. Prove that, if in a plane the ratio of the distances from two
points be the same for each of three points A, B, and C, the two points are
inverse points with regard to the circle ABC. Prove also that the line
bisecting BC at right angles meets the lines BA and CA in two such points.
Ex. 656. If a circle S touch the circumcircle of a triangle ABC at P,
prove that the tangents to S from A, B, C are in the ratios AP : BP : CP.
What does this result become when the radius of the circle S increases
indefinitely ?
Ex. 657. PQ and RS are interior and exterior common tangents to two
circles. The circles QSR and SRP cut PQ atp, q respectively; and the
circles PQS, PQR cut RS at r, s respectively. Shew that circles will pass
through Q, S, q, s and through P, R, p, r, and that the rectangle contained
by their radii equals the rectangle contained by the radii of the original
circles.
Ex. 658. A triangle of given shape is inscribed on a given triangle.
Shew that the locus of its centroid is in general six straight lines.
Ex. 659. A circle U of constant radius is described, having its centre at
any point of the circumference of a fixed circle whose centre is O ; the
variable circle. U cuts another fixed circle V ; Y is the foot of the per-
pendicular from O on the common chord of U and V. Prove that the locus
of Y is a circle.
Ex. 660. If two fixed circles be cut by a variable straight line in four
points in a harmonic range, show that the product of the perpendiculars
upon it from the centres of the circles is constant.
Ex. 661. Through any point O in the plane of a triangle ABC is drawn
a transversal, cutting the sides in P, Q, R. The lines OA, OB, OC are
bisected in A', B', C ; and the segments QR, RP, PQ of the transversal
are bisected in P', Q', R'.
Show that the three lines A'P', B'Q', C'R' are concurrent.
Ex. 662. The four points A BCD form a quadrilateral of which the
diagonals AC, BD intersect in O, and A', B', C, D' are the inverse points
with regard to O as origin of A, B, C, D respectively. Show that A'B'C'D'
is a quadrilateral having its angles supplementary to those of A BCD and
that, if turned over, it may be placed in the plane so as to have sides and
diagonals parallel to those of A BCD.
158 MISCELLANEOUS EXERCISES
Ex. 663. If from any point on the circumference of a circle perpen-
diculars are drawn to the four sides and to the diagonals of an inscribed
quadrilateral, prove that the rectangle contained by the perpendiculars on
either pair of opposite sides is equal to that contained by the perpendiculars
on the diagonals.
Ex. 664. If a system of circles be drawn so that each bisects the cir-
cumferences of two given circles, then the polars of a given point with
respect to the system of circles will be concurrent.
Ex. 665. A line is drawn cutting two non-intersecting circles ; find a
construction determining two points on this line such that each is the
point of intersection of the polars of the other point with respect to the
two circles.
Ex. 666. If, on the sides BC, CD of a quadrilateral ABCD of which
two opposite angles at B and D are equal (the other two opposite angles
being unequal) points E and F be taken such that the areas of the triangles
AED, AFB are equal, prove that the radical axis of the circles on BF, ED
as diameters passes through A.
Ex. 667. Two opposite sides of a quadrilateral inscribable in a circle
lie along two given lines OX, OY and the intersection of the diagonals is
given; show that the locus of the centres of the circles is a straight line.
Ex. 668. Two circles intersect orthogonally at a point P, and O is any
point on any circle which touches the two former circles at Q and Q'.
Show that the angle of intersection of the circumcircles of the triangles
OPQ, OPQ' is half a right angle.
Ex. 669. The triangles AiBjCi, A0B2C2 are reciprocal with respect to a
given circle; BgCg, C^A^ intersect in P^ and B^Cj, C^/K^ in Pg. Show that
the radical axis of the circles which circumscribe the triangles P^A^Bg,
PgAgBj^ passes through the centre of the given circle.
Ex. 670. Show that if each of two pairs of opposite vertices of a quadri-
lateral is conjugate with regard to a circle the third pair is also ; and that
the circle is one of a coaxal system of which the line of coUinearity of the
middle points of the diagonals is the radical axis.
intersection of Cg and C., passes through the centre of C^, and the chord
of intersection of C3 and C^ through the centre of C^ ; show that the chord
of intersection of C^ and C.^ passes through the centre of C3.
MISCELLANEOUS EXERCISES 159
Ex. 672. A system of spheres touch a plane P (on either side of the
plane) at a point O. A plane Q, not passing through O, cuts P in the line I,
touches two of the spheres in L^ and L2 respectively, and cuts the other
spheres. Show that the system of circles in which Q cuts the spheres is
coaxal, with L^ and L2 as limiting points and I as radical axis.
Ex. 673. Show that the locus of a point at which two given portions of
the same straight line subtend equal angles is a circle.
Ex. 674. Two variable circles touch each of two fixed circles and each
other ; show that the locus of the point of contact of the variable circles is
a circle.
Ex. 675. A, B, C, D are four circles in a plane, each being external to
the other three and touching two of them. Show that the four points of
contact are concyclic.
Ex. 676. Three circles meet in a point O. The common chord of the
first and second passes through the centre of the third, and the common
chord of the first and third passes through the centre of the second. Prove
by inversion with respect to O that the common chord of the second and
the third passes through the centre of the first.
Ex. 677. AOB is a right-angled triangle, O is the right angle, and OL
is the perpendicular to AB. On the other side of OB remote from A the
square OBGF is described, and the line AG cuts OL in M. Prove that
OM~AB"^OL*
Ex. 678. If A, B are conjugate points with respect to a circle (see note
to Ex. 899), then the tangent to the circle from O, the mid-point of AB, is
equal to OA.
Ex. 679. The sides BC, DA of the quadrilateral ABCD are cut by any
line in the points K, L respectively. AC, BD meet in X; AK, BL meet in
Y; CL, DK meet in Z and BC, AD meet in E. Prove that
X {KCZD} = {EALD} = X {KCYD},
and that XYZ is a straight line.
INDEX.
Angles of intersection of curves 76
Anharmonic ratio 123
Apollonius' circle 78
Apollonius' theorem 20
Axis, radical 87
Base of range 123
Centre of inversion 100
Centre of similitude 73
Centre, radical 90
Centroid 11
Centroid of triangle 30
Ceva 46
Chord of contact 62
Circle of Apollonius 78
Circle of infinite radius 10, 69
Circle of inversion 100
Circum-centre 22
Circum-circle 22
Coaxal circles 87
Collinear 22
Complete quadrangle 138
Complete quadrilateral 138
Concurrent 22
Conjugates, harmonic 53
Constant of inversion 100
Contact problems 83
Cross-ratio 54
Cross-ratio of pencil 125
Cross-ratio of pencil of parallel lines
129
Cross-ratio of range 123
Cross-ratios and projection 133
Theorem 146
Diagonal of quadrilateral 138
Diagonal-point of quadrangle 138
Duality 136
Ellipse 119
Equicross 123
Escribed circle 24
Ex-centre 24
Ex-circle 24
Figures in perspective 146
Harmonic conjugates 53
Harmonic pencil 58
Harmonic progression 54
Harmonic range 53, 141
Harmonic section 53
In-centre 23
In-circle 23
Infinity 6
Inverse points 100
162
INDEX
Inversion 100
Inversion, centre of 100
Inversion, circle of 100
Inversion, constant of 100
Inversion, radius of 100
Join of points 137
Pole 62, 63
Principle of duality 136
Projection 114
Projective construction for
harmonic 141
Ptolemy 80
Ptolemy's Theorem 80
fourth
Limiting points 94
Line at infinity 9
Quadrangle 138
Quadrilateral 138
Medial triangle 29
Median 29
Meet of lines 137
Menelaus 49
Kadieal axis 87
Kadical centre 90
Kadius of inversion
Bangs 123
100
Nine-points centre 36
Nine-points circle 35
Notation for triangle 16
Orthocentre 31
Orthogonal circles 76
Orthogonal projection 114
Parallel translation 84
Peaucellier's Cell 104
Pedal triangle 32
Pencil 56, 123
Pencil, harmonic 58
Perspective 146
Point at infinity 7
Points, limiting 94
Polar 62, 63
Polar circle 143
Salmon's theorem 70
Self-conjugate triangle 143
Self-polar triangle 143
Sense of a line 1
Sense of an angle 5
Similarly situated 72
Similitude 71
Similitude, centre of 73
Simson line 37
Theorem of Apollonius 20
Theorem of Ceva 46
Theorem of Menelaus 49
Transversal of pencil 56
Triangle 16
Vertex of pencil 56, 123
CAMBRIDGE : PRINTED BY JOHN CLAY, M.A. AT THE UNIVERSITY PRESS.
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